NCERT Solution Class 11 Physics Chapter 4 Laws of Motion- Param Himalaya
4.1 Give the magnitude and direction of the net force acting on
(a) a drop of rain falling down with a constant speed,
(b) a cork of mass 10 g floating on water,
(c) a kite skillfully held stationary in the sky,
(d) a car moving with a constant velocity of 30 km/h on a rough road,
(e) a high-speed electron in space far from all material objects, and free of electric and magnetic fields.
Solution:
(a) Raindrop falling with constant speed:
* Magnitude: 0 N
* Direction: No direction (since the net force is zero)
When an object moves with constant speed, its acceleration is zero. According to Newton's second law (F = ma), if acceleration is zero, the net force acting on the object is also zero.
(b) Cork floating on water:
* Magnitude: 0 N
* Direction: No direction (since the net force is zero)
The cork is in equilibrium, meaning the upward buoyant force from the water exactly balances the downward force of gravity. Therefore, the net force is zero.
(c) Kite held stationary in the sky:
* Magnitude: Non-zero (depends on wind speed and string tension)
* Direction: Varies with wind direction and string tension. The net force must be equal and opposite to the force exerted by the wind on the kite, keeping it stationary.
(d) Car moving with constant velocity:
* Magnitude: 0 N
* Direction: No direction (since the net force is zero)
Similar to the raindrop, a car moving with constant velocity has zero acceleration, and thus zero net force. The forces acting on the car, such as friction and air resistance, are balanced by the engine's force.
(e) High-speed electron in space:
* Magnitude: 0 N
* Direction: No direction (since the net force is zero)
In the absence of external forces like gravity, electric fields, or magnetic fields, the electron will continue to move in a straight line with constant velocity. This implies zero acceleration and zero net force.
4.2 A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction
and magnitude of the net force on the pebble,
(a) during its upward motion,
(b) during its downward motion,
(c) at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction? Ignore air resistance.
Solution:
(a) During upward motion:
* Direction: Downward
* Magnitude: 0.49 N
The only force acting on the pebble is gravity, which pulls it downward. The magnitude of this force can be calculated using Newton's second law:
F = ma = (0.05 kg)(9.8 m/s²) = 0.49 N
(b) During downward motion:
* Direction: Downward
* Magnitude: 0.49 N
Again, the only force acting on the pebble is gravity, which remains constant in magnitude and direction throughout the motion.
(c) At the highest point:
* Direction: Downward
* Magnitude: 0.49 N
Even though the pebble is momentarily at rest at the highest point, gravity continues to act on it, pulling it downward with a force of 0.49 N.
If the pebble was thrown at an angle of 45°:
The direction of the net force (gravity) would still be downward, and its magnitude would remain the same (0.49 N) at all points of the motion. The angle of projection only affects the trajectory of the pebble, not the net force acting on it.
4.3 Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg,
(a) just after it is dropped from the window of a stationary train,
(b) just after it is dropped from the window of a train running at a constant
velocity of 36 km/h,
(c ) just after it is dropped from the window of a train accelerating with 1 m s-2
(d) lying on the floor of a train which is accelerating with 1 m s-2, the stone being at rest relative to the train.Neglect air resistance throughout.
Solution:
(a) Just after it is dropped from the window of a stationary train:
* Magnitude: 0.98 N
* Direction: Downward
The only force acting on the stone is gravity, which pulls it downward. The magnitude of this force can be calculated using Newton's second law:
F = ma = (0.1 kg)(9.8 m/s²) = 0.98 N
(b) Just after it is dropped from the window of a train running at a constant velocity of 36 km/h:
* Magnitude: 0.98 N
* Direction: Downward
Even though the train is moving, the stone's horizontal motion is independent of its vertical motion. The only force acting on the stone after it's dropped is gravity, pulling it downward with a force of 0.98 N.
(c) Just after it is dropped from the window of a train accelerating with 1 m/s²:
* Magnitude: 0.98 N
* Direction: Downward
Similar to the previous case, the stone's horizontal motion is independent of its vertical motion. The only force acting on the stone after it's dropped is gravity, pulling it downward with a force of 0.98 N.
(d) Lying on the floor of a train which is accelerating with 1 m/s², the stone being at rest relative to the train:
* Magnitude: 0.1 N
* Direction: Forward (in the direction of the train's acceleration)
In this case, the stone is accelerating horizontally along with the train. This acceleration requires a horizontal force to act on the stone. This force is provided by the friction between the stone and the floor of the train. The magnitude of this force can be calculated using Newton's second law:
F = ma = (0.1 kg)(1 m/s²) = 0.1 N
4.4 One end of a string of length $l$ is connected to a particle of mass $m$ and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed $v$, the net force on the particle (directed towards the centre) is:
(i) $T$,
(ii) $T - \frac{mv^2}{l}$,
(iii) $T + \frac{mv^2}{l}$,
(iv) $0$
$T$ is the tension in the string. [Choose the correct alternative.]
Solution:
In this scenario, the particle is undergoing uniform circular motion. The only force acting on the particle towards the center of the circle is the tension in the string, T.
Therefore, the net force on the particle is (i) T.
Explanation:
* Centripetal Force: For an object to move in a circle, there must be a force acting towards the center of the circle, known as the centripetal force.
* In this case: The tension in the string provides the necessary centripetal force to keep the particle moving in a circular path.
* No other forces: There are no other forces acting on the particle in the horizontal plane.
Hence, the net force is solely the tension, T.
4.5 A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m/s. How long does the body take to stop ?
Solution:
Here
Force (F) = -50N (retarding force)
Mass m = 20 kg
Final speed (v) = 0
Initial speed (u) = 15 m/s
So
$$F = ma$$
$$a = \frac{F}{m}$$
$$a= - \frac{50}{20}$$
$$a = -2.5 m/s^{2}$$
Using the equation
$$v = u+at$$
$$0 = 15 + (-2.5)t$$
$$2.5t = 15$$
$$t = \frac{15}{2.5}$$
$$t = 6 s$$
4.6 A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m/s to 3.5 m s-1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force ?
Solution
Given:
* Mass (m) = 3.0 kg
* Initial velocity (u) = 2.0 m/s
* Final velocity (v) = 3.5 m/s
* Time (t) = 25 s
To find:
* Magnitude of the force (F)
* Direction of the force
Steps:
* Calculate the acceleration:
* Using the first equation of motion:
* v = u + at
* a = (v - u) / t
* Calculate the force:
* Using Newton's second law of motion:
* F = ma
Calculations:
* Acceleration:
* a = (3.5 - 2.0) / 25
* a = 0.06 m/s²
* Force:
* F = 3.0 * 0.06
* F = 0.18 N
Conclusion:
The magnitude of the force is 0.18 N. Since the direction of the motion of the body remains unchanged, the direction of the force is in the same direction as the motion.
Therefore, the force has a magnitude of 0.18 N and is in the direction of the body's motion.
4.7 A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.
Solution:
Understanding the Problem
* We have two forces acting perpendicularly on a body.
* We need to find the net force and then use Newton's second law to find the acceleration.
Steps
* Find the net force:
* Since the forces are perpendicular,then resultant force given by :
$R = \sqrt{A^{2}+B^{2}+2ABcos \theta}$
$F = \sqrt{8^{2}+6^{2}+2 \times 8 \times 6 cos90^\circ}$
$F = \sqrt{8^{2}+6^{2}+2 \times 8 \times 6 \times 0}$
$F = \sqrt{8^{2}+6^{2} }$
$F= \sqrt{64+36}$
$F= \sqrt{100}$
$F= 10N$
* Find the acceleration:
* Use Newton's second law: F = ma
$a = \frac{F}{m}$
$a = \frac{10 N}{5 kg}$
$a= 2 m/s^{2}$
* Find the direction of acceleration:
* The direction of the acceleration is the same as the direction of the net force.
* To find the angle, use trigonometry:
$tan \theta= \frac{6 N}{8 N}$
$tan \theta= 0.75$
θ = tan⁻¹(0.75) ≈ 36.87°
Conclusion
The acceleration of the body is 2 m/s² in a direction 36.87° from the 8 N force.
4.8 The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle ? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg.
Solution:
Steps to solve:
1. Convert the initial velocity from km/h to m/s:
$u = 36 \text{ km/h}$
$u= 36 \cdot \frac{1000 \text{ m}}{3600 \text{ s}}$
$u= 10 \text{ m/s}$
2. Calculate the acceleration of the vehicle:
$a = \frac{v - u}{t}$
$a= \frac{0 - 10 \text{ m/s}}{4 \text{ s}}$
$a= -2.5 \text{ m/s}^2$
3. Calculate the total mass of the vehicle and driver:
$m = m_{\text{vehicle}} + m_{\text{driver}} = 400 \text{ kg} + 65 \text{ kg}$
$m= 465 \text{ kg}$
4. Calculate the average retarding force on the vehicle:
$F = ma = 465 \text{ kg} \cdot (-2.5 \text{m/s}^2)$
$F= -1162.5 \text{ N}$
Answer:
The average retarding force on the vehicle is -1162.5 N. The negative sign indicates that the force is acting in the opposite direction of the motion of the vehicle.
4.9 A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m s-2. Calculate the initial thrust (force) of the blast.
Solution:
Steps to solve:
1. Calculate the weight of the rocket:
$W = mg$
$W= 20000 \text{ kg} \cdot 9.8 \text{ m/s}^2$
$W= 196000 \text{ N}$
2. Apply Newton's second law of motion:
$F_{\text{net}} = F_{\text{thrust}} - W = ma$
3. Solve for the thrust force:
$F_{\text{thrust}} = ma + W$
$F_{\text{thrust}} = 20000 \text{ kg} \cdot 5.0 \text{ m/s}^2 + 196000 \text{ N}$
$F=100000 \text{ N} + 196000 \text{ N}$
$F= 296000 \text{ N}$
Answer:
The initial thrust of the blast is 296000 N.
4.10 A body of mass 0.40 kg moving initially with a constant speed of 10 m s-1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its position at t = –5 s, 25 s, 100 s.
Solution:
Acceleration:
Calculate the acceleration using Newton's second law: F = ma
$a = \frac{F}{m}$
$a= \frac{-8.0 N}{0.40 kg}$
$a= -20 m/s^2 (negative sign indicates southward direction)$
Velocity:
Use the equation v = u + at to find the velocity at any time t:
$v = 10 m/s - 20 m/s^2 * t$
Displacement:
Use the equation $s = ut + 0.5at^2$ to find the displacement at any time t:
* s = 10 m/s * t - 0.5 * 20 m/s^2 * t^2
Now, let's calculate the positions at the given times:
t = -5 s:
* This time is before the force is applied. The body moves with a constant velocity of 10 m/s north south $a = 0 m/s^2$
Displacement
$s= 10 m/s \times (-5 s) = -50 m$(50 m south)
t = 25 s:
The force has been applied for 25 seconds.
* Velocity = 10 m/s - 20 m/s^2 * 25 s = -490 m/s (southward)
* Displacement = 10 m/s * 25 s - 0.5 * 20 m/s^2 * (25 s)^2 = -6000 m (6000 m south)
t = 100 s:
* The force has been applied for 100 seconds.
* Velocity = 10 m/s - 20 m/s^2 * 100 s = -1990 m/s (southward)
* Displacement = 10 m/s * 100 s - 0.5 * 20 m/s^2 * (100 s)^2 = -99000 m (99000 m south)
Therefore, the positions of the body at t = -5 s, 25 s, and 100 s are -50 m, -6000 m, and -99000 m, respectively. All positions are south of the s
tarting point (x = 0).
4.11 A truck starts from rest and accelerates uniformly at 2.0 m s-2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t =11s ? (Neglect air resistance.)
Solution:
(a) Velocity of the stone at t = 11 s:
* The stone inherits the horizontal velocity of the truck at the moment it is dropped.
* The horizontal velocity of the truck at t = 10 s can be calculated using the formula: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
* Here, u = 0 m/s (truck starts from rest), a = 2 m/s², and t = 10 s.
* So, v = 0 + 2 * 10 = 20 m/s.
* This is the horizontal velocity of the stone at t = 11 s.
* The stone also has a vertical velocity due to gravity. The vertical velocity at t = 11 s can be calculated using the same formula, but with the acceleration due to gravity (g = 9.81 m/s²).
* The time of fall is 1 second (from t = 10 s to t = 11 s).
* So, vertical velocity = 0 + 9.81 * 1 = 9.81 m/s.
* The resultant velocity of the stone is the vector sum of the horizontal and vertical velocities.
* Using Pythagoras theorem:
Resultant velocity = sqrt(horizontal velocity² + vertical velocity²)
= sqrt(20² + 9.81²)
= 22.28 m/s (approximately)
Therefore, the velocity of the stone at t = 11 s is approximately 22.28 m/s.
(b) Acceleration of the stone at t = 11 s:
Once the stone is dropped, the only force acting on it is gravity. Therefore, the acceleration of the stone is the acceleration due to gravity, which is approximately 9.81 m/s² downwards.
4.12 A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 m s-1.What is the trajectory of the bob if the string is cut when the bob is (a) at one of its extreme positions, (b) at its mean position.
Solution:
(a) When the bob is at one of its extreme positions, its velocity is momentarily zero. If the string is cut at this instant, the only force acting on the bob is gravity. Therefore, the bob will fall vertically downwards under the influence of gravity.
(b) At the mean position, the bob has a horizontal velocity of 1 m/s. If the string is cut at this point, the bob will behave like a projectile with an initial horizontal velocity. Under the combined action of gravity and its initial horizontal velocity, the bob will follow a parabolic path.
4.13 A man of mass 70 kg stands on a weighing scale in a lift which is moving
(a) upwards with a uniform speed of 10 m s-1
(b) downwards with a uniform acceleration of 5 m s-2
(c) upwards with a uniform acceleration of 5 m s-2
What would be the readings on the scale in each case?
(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity ?
Solution:
Understanding the Problem:
In this problem, we're asked to determine the apparent weight of a person in a lift under various conditions. The apparent weight is the force exerted by the person on the weighing scale, and it can be different from the actual weight due to the acceleration of the lift.
Key Concept: Newton's Second Law of Motion:
Newton's Second Law states that the net force acting on an object is equal to the product of its mass and acceleration. In this case, the net force acting on the person is the difference between the normal force (N) exerted by the scale and the gravitational force (mg).
Solution:
a) Upwards with a uniform speed of 10 m/s:
* Since the lift is moving with a constant speed, there is no acceleration.
* Therefore, the net force on the person is zero.
* The normal force (N) equals the gravitational force (mg).
* Reading on the scale = N = mg = 70 kg × 9.8 m/s² = 686 N
b) Downwards with a uniform acceleration of 5 m/s²:
* The net force on the person is downward.
* N - mg = ma (where a is the acceleration of the lift)
* N = m(g + a) = 70 kg × (9.8 m/s² + 5 m/s²) = 1036 N
c) Upwards with a uniform acceleration of 5 m/s²:
* The net force on the person is upward.
* N - mg = ma
* N = m(g - a) = 70 kg × (9.8 m/s² - 5 m/s²) = 336 N
d) Freely falling under gravity:
* In this case, the lift and the person inside are both accelerating downwards with the acceleration due to gravity (g).
* The normal force (N) is zero, as the person is not in contact with the scale.
* Reading on the scale = 0 N
Conclusion:
The readings on the scale in each case are:
* a) 686 N
* b) 1036 N
* c) 336 N
* d) 0 N
4.14 Figure 4.16 shows the position-time graph of a particle of mass 4 kg. What is
the (a) force on the particle for t < 0, t > 4 s, 0 < t < 4 s?
(b) impulse at t = 0 and t = 4 s ? (Consider one-dimensional motion only).
Solution:
(a) Force on the particle:
* t < 0: The graph shows that the particle is at rest. Therefore, the net force on it is zero.
* 0 < t < 4 s: The graph shows a straight line with a constant slope, indicating a constant velocity. This means the acceleration is zero. Since force is mass times acceleration, the force on the particle is zero during this interval as well.
* t > 4 s: The graph shows that the particle is at rest again. Therefore, the net force on it is zero.
(b) Impulse at t = 0 and t = 4 s:
Impulse is the change in momentum. Since momentum is mass times velocity, and the velocity doesn't change at t = 0 or t = 4 s (the particle is at rest at these times), the change in momentum is zero. Therefore, the impulse at both t = 0 and t = 4 s is zero.
Summary:
* Force: Zero for all time intervals (t < 0, 0 < t < 4 s, t > 4 s)
* Impulse: Zero at t = 0 and t = 4 s
*
4.15 Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force F = 600 N is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case?
Solution :
Analyzing the Tension in the String
Understanding the Problem:
Two masses are connected by a string and pulled by a force. We need to find the tension in the string for different points of application of the force.
Key Points:
* The string is assumed to be massless and inextensible.
* The surface is smooth, meaning there is no friction.
Analyzing the Cases:
(i) Force Applied to Mass A
Free Body Diagram for Mass A: The force F acts on A to the right, and the tension T acts on A to the left.
Newton's Second Law for Mass A:
$F - T = m_{1}a$ ....... (1)
Free Body Diagram for Mass B: The tension T acts on B to the right.
Newton's Second Law for Mass B:
$T = m_{2}a$ ......... (2)
Since the string is inextensible, both masses will accelerate at the same rate (a).
Combining the two equations:
$F = (m_{1}+ m_{2})a$
Solving for a:
$a = \frac{F}{(m_{1}+ m_{2})}$
$a= \frac{600 N}{(10 kg + 20 kg)}$
$a = 20 m/s^{2}$
Substituting a into the equation for T:
$T = m_{2}a$
$T= 20 \times 20$
$T= 400 N$
(ii) Force Applied to Mass B
* Free Body Diagram for Mass A: The tension T acts on A to the right.
* Newton's Second Law for Mass A:
$T = m_{1}a$
Free Body Diagram for Mass B: The force F acts on B to the right, and the tension T acts on B to the left.
Newton's Second Law for Mass B:
$F - T = m_{2}a$
Again, the acceleration is the same for both masses.
Combining the equations:
$F = (m_{1} + m_{2})a$
Solving for a:
$a = \frac{F}{(mA + mB)}$
$a= \frac{600 N}{(10 kg + 20 kg)}$
$a = 20 m/s^{2}$ (same as before)
Substituting a into the equation for T:
$T = m_{1}a$
$T =10 \times 20$
$T= 200 N$
Conclusion:
* The tension in the string is 400 N when the force is applied to mass A.
* The tension in the string is 200 N when the force is applied to mass B.
Note: In both cases, the acceleration of the system remains the same. However, the distribution of tension between the two masses changes depending on the point of application of the force.
4.16 Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.
Solution:
According to the question,
$m_{1}= 8 kg$
$m_{2} = 12 kg$
Tension = T
The heavier mass $m_{2}$ will move downwards and the smaller mass $m_{1}$ will move upwards.
On Applying Newton’s second law,
For mass $m_{1}$:
$T – m_{1}.g = m_{1}.a $ —– (1)
For mass m2:
$m_{2}.g – T = m_{2}.a$ ——(2)
Add (1) and (2)
$T – m_{1}.g + m_{2}.g – T = m_{1}.a + m_{2}.a$
$m_{2}.g – m_{1}.g = m_{1}.a + m_{2}.a$
$(m2 – m1) g = (m1 + m2) a$
$(m1 + m2) a= (m2 – m1)g$
$a = \frac{(m2 – m1)}{(m1 + m2)}.g$. --- 3
$a= \frac{[ (12 – 8)}{(12 + 8)]} \times 10$
$a= \frac{(4)}{(20)}\times 10$
$a= 2m/s$
Therefore, acceleration of the mass is $2 m/s^{2}$
From equation -2
$m_{2}.g – T = m_{2}.a$
$T = m_{2}.g - m_{2}.a$
$T = 12 \times 10 - 12 \times 2$
$T = 120 - 24$
$T = 96N$
Hence, the tension on the string is 96 N
4.17 A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.
Solution:
Conservation of Momentum:
In the absence of external forces, the total momentum of a system remains constant.
Initial Momentum:
Since the nucleus is at rest, its initial momentum is zero.
Final Momentum:
After the disintegration, the nucleus breaks down into two smaller nuclei. Let's denote their masses as m1 and m2, and their velocities as v1 and v2, respectively.
The final momentum of the system is given by:
$m_{1}v_{1} + m_{2}v_{2}$
Conservation of Momentum Equation:
According to the conservation of momentum, the initial momentum must equal the final momentum. Therefore, we have:
$0 = m_{1}v_{1} + m_{2}v_{2}$
Solving for Velocities:
Rearranging the equation, we get:
$m_{1}v_{1}= -m_{2}v_{2}$
$v_{1} = \frac{ -m_{2}v_{2}}{m_{1}}$
This equation tells us that the product of the mass and velocity of one nucleus is equal in magnitude but opposite in sign to the product of the mass and velocity of the other nucleus. This implies that the two nuclei must move in opposite directions to conserve momentum.
Conclusion:
Therefore, if a nucleus at rest disintegrates into two smaller nuclei, the products must move in opposite directions to satisfy the conservation of momentum.
4.18 Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 m/s.collide and rebound with the same speed. What is the impulse imparted to each ball due to the other ?
Solution:
* Initial momentum of each ball:
Mass (m) = 0.05 kg
Velocity (v) = 6 m/s
Momentum (p) = m * v
p= 0.05 kg * 6 m/s
p = 0.3 kg m/s
* Final momentum of each ball:
After the collision, the velocity of each ball reverses direction but remains the same magnitude.
Therefore, the final momentum of each ball is -0.3 kg m/s (negative sign indicates the opposite direction).
* Impulse:
Impulse (J) = Change in momentum (Δp)
J= Final momentum - Initial momentum
* For each ball, J = -0.3 kg m/s - 0.3 kg m/s
J = -0.6 kg m/s
Therefore, the impulse imparted to each ball due to the other is -0.6 kg m/s. The negative sign indicates that the impulse is in the opposite direction of the initial velocity.
4.19 A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m/s, what is the recoil speed of the gun ?
Solution:
Understanding the Problem
We're given:
* Mass of the shell (m): 0.020 kg
* Mass of the gun (M): 100 kg
* Muzzle speed of the shell (v): 80 m/s
* We need to find the recoil speed of the gun (V).
According to the Law of Conservation of Momentum, the total momentum of a system remains constant if no external forces act on it. In this case, the system is the gun and the shell.
Before firing, both the gun and the shell are at rest, so their total momentum is 0. After firing, the gun recoils in the opposite direction to the shell.
Therefore, we can write:
* Initial momentum = Final momentum
$m_{1}u_{1}+ m_{2}u_{2} = m_{1}v_{1}+ m_{2}v_{2}$
$0 = mv - MV$
Where:
* mv is the momentum of the shell
* MV is the momentum of the gun
Solving for the Recoil Speed
Rearranging the equation to solve for V:
* V = mv / M
Substituting the given values:
* V = (0.020 kg)(80 m/s) / (100 kg)
* V = 0.016 m/s
Therefore, the recoil speed of the gun is 0.016 m/s.
4.20 A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball ? (Mass of the ball is 0.15 kg.)
Solution:
Here's how we can calculate the impulse imparted to the ball:
1. Convert the initial speed to m/s:
initial_speed = 54 km/h * (1000 m/km) / (3600 s/h) = 15 m/s
2. Calculate the initial momentum of the ball:
initial_momentum = mass * initial_speed = 0.15 kg * 15 m/s = 2.25 kg m/s
3. Calculate the final momentum of the ball:
The ball is deflected by 45 degrees, so the change in the x-component of the momentum is:
Δpx = final_momentum_x - initial_momentum_x = 2.25 * cos(45°) - (-2.25 * cos(45°)) = 3.18 kg m/s
4. Calculate the impulse:
Impulse is equal to the change in momentum:
impulse = Δpx = 3.18 kg m/s
Therefore, the impulse imparted to the ball is 3.18 kg m/s.
4.21 A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev/min in a horizontal plane. What is the tension in the string ? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N ?
Solution:
Understanding the Problem
When a stone is whirled in a circle, the tension in the string provides the necessary centripetal force to keep the stone moving in a circular path.
Given:
* Mass of the stone (m) = 0.25 kg
* Radius of the circle (r) = 1.5 m
* Angular speed (ω) = 40 rev/min
* Maximum tension the string can withstand (T_max) = 200 N
Solution:
Part 1: Tension in the string
First, we need to convert the angular speed from rev/min to rad/s:
ω = 40 rev/min * (2π rad/rev) * (1 min/60 s) = 4π/3 rad/s
The centripetal force (F_c) required to keep the stone moving in a circle is given by:
F_c = mω²r
Since the tension in the string provides this centripetal force:
T = F_c = mω²r
Substituting the given values:
T = (0.25 kg) * (4π/3 rad/s)² * (1.5 m) ≈ 6.58 N
Part 2: Maximum speed
We can rearrange the centripetal force equation to find the maximum linear speed (v_max) that the stone can have:
v_max = √(T_max * r / m)
Substituting the given values:
v_max = √(200 N * 1.5 m / 0.25 kg) ≈ 34.64 m/s
Therefore, the tension in the string is approximately 6.58 N, and the maximum speed the stone can be whirled around is about 34.64 m/s.
4.22 If, in Exercise 4.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks :
(a) the stone moves radially outwards,
(b) the stone flies off tangentially from the instant the string breaks,
(c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle ?
Solution:
Option (b) is correct. When the string breaks, the stone will move in the direction of the velocity at that instant. Now, the direction of velocity vector is tangential to the path of the stone at that instant. Hence, the stone will fly off tangentially from the instant the string breaks.
4.23 Explain why
(a) a horse cannot pull a cart and run in empty space,
(b) passengers are thrown forward from their seats when a speeding bus stops
suddenly,
(c) it is easier to pull a lawn mower than to push it,
(d) a cricketer moves his hands backwards while holding a catch.
Solution :
(a) A horse cannot pull a cart and run in empty space because it needs a surface to push against to generate the reaction force that propels it forward. This is in accordance with Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction. In empty space, there is no surface to push against, so the horse cannot generate the necessary reaction force to move itself and the cart.
(b) Passengers are thrown forward from their seats when a speeding bus stops suddenly due to inertia. Inertia is the tendency of an object to resist changes in its motion. When the bus is moving at a constant speed, the passengers inside are also moving at the same speed. However, when the bus stops suddenly, the passengers' bodies tend to continue moving forward due to their inertia. This causes them to be thrown forward from their seats.
(c) It is easier to pull a lawn mower than to push it because when pulling, the downward force of gravity on the handles helps to increase the friction between the wheels and the ground, making it easier to move the mower. When pushing, the upward force of gravity on the handles reduces the friction, making it more difficult to move the mower.
(d) A cricketer moves his hands backwards while holding a catch to increase the time taken to stop the ball. This reduces the force exerted on the hands, preventing injury. This is in accordance with the impulse-momentum theorem, which states that the impulse (force multiplied by time) is equal to the change in momentum. By increasing the time taken to stop the ball, the force exerted on the hands is reduced.