12. A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because
(a) the batsman did not hit the ball hard enough.
(b) velocity is proportional to the force exerted on the ball.
(c) there is a force on the ball opposing the motion.
(d) there is no unbalanced force on the ball, so the ball would want to come to rest.
Solution:
When the ball rolls on the flat surface of the ground, its motion is opposed by the force of friction (the friction arises between the ground and the ball). This frictional force eventually stops the ball. Therefore, the correct answer is (c).
If the surface of the level ground is lubricated (with oil or some other lubricant), the friction that arises between the ball and the ground will reduce, which will enable the ball to roll for a longer distance.
13. A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if it’s mass is 7 tonnes (Hint: 1 tonne = 1000 kg.)
Solution:
Given, distance covered by the truck (s) = 400 meters
Time taken to cover the distance (t) = 20 seconds
The initial velocity of the truck (u) = 0 (since it starts from a state of rest)
From the equations of motion, s = ut + 1/2at2
Therefore, 400 = 0(20s) + 1/2(a) (400s2) = 2ms-2
The acceleration of the truck is equal to 2 ms-2
As per the second law of motion, Force = Mass x Acceleration
Mass of the truck = 7 tonnes = 7000kg
Force acting on the truck = 7000kg x 2m. s-2 = 14000 kg.m.s2 = 14000 N
Therefore, a force of 14000 N is acting on the truck.
14. A stone of 1 kg is thrown with a velocity of 20 ms-1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?
Solution:
Given, Mass of the stone (m) = 1kg
Initial velocity (u) = 20m/s
Terminal velocity (v) = 0 m/s (the stone reaches a position of rest)
Distance travelled by the stone (s) = 50 m
As per the third equation of motion
v2 = u2 + 2as
Substituting the values in the above equation we get,
02 = (20)2 + 2(a)(50)
-400 = 100a
a = -400/100 = -4m/s2 (retardation)
We know that
F = m × a
Substituting above obtained value of a = -4 in F = m x a
We get,
F = 1 × (-4) = -4N
Here the negative sign indicates the opposing force which is Friction.
15. An 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate: (a) the net accelerating force and (b) the acceleration of the train
Solution:
(a) Given, the force exerted by the train (F) = 40,000 N
Force of friction = -5000 N (the negative sign indicates that the force is applied in the opposite direction)
Therefore, the net accelerating force = sum of all forces = 40,000 N + (-5000 N) = 35,000 N
(b) Total mass of the train = mass of engine + mass of each wagon = 8000kg + 5 × 2000kg
The total mass of the train is 18000 kg.
As per the second law of motion, F = ma (or: a = F/m)
Therefore, acceleration of the train = (net accelerating force) / (total mass of the train)
= 35,000/18,000 = 1.94 ms-2
The acceleration of the train is 1.94 m.s-2.
16. An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 ms-2?
Solution:
Given, mass of the vehicle (m) = 1500 kg
Acceleration (a) = -1.7 ms-2
As per the second law of motion, F = ma
F = 1500kg × (-1.7 ms-2) = -2550 N
Hence, the force between the automobile and the road is -2550 N, in the opposite direction of the automobile’s motion.
17. What is the momentum of an object of mass m, moving with a velocity v?
(a) (mv)2
(b) mv2
(c) 1/2 mv2
(d) mv
Solution:
The momentum of an object is defined as the product of its mass m and velocity v
Momentum = mass x velocity
Hence, the correct answer is mv i.e option (d).
18. Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?
Solution:
Since the velocity of the cabinet is constant, its acceleration must be zero. Therefore, the effective force acting on it is also zero. This implies that the magnitude of opposing frictional force is equal to the force exerted on the cabinet, which is 200 N. Therefore, the total friction force is -200 N.
19. Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 ms-1 before the collision during which they stick together. What will be the velocity of the combined object after collision?
Solution:
Given
Mass of first object, m1 = 1.5 kg
Mass of second object, m2 = 1.5 kg
Velocity of first object before collision, v1 = 2.5 m/s
The velocity of the second object which is moving in the opposite direction, v2 = -2.5 m/s
We know that,
Total momentum before collision = Total momentum after collision
m1v1 + m2v2 = (m1 + m2)v
1.5(2.5) + 1.5 (-2.5) = (1.5 + 1.5)v
3.75 – 3.75 = 3v
v = 0
Therefore, the velocity of the combined object after the collision is 0 m/s.
20. According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.
Solution:
Since the truck has a very high mass, the static friction between the road and the truck is high. When pushing the truck with a small force, the frictional force cancels out the applied force and the truck does not move. This implies that the two forces are equal in magnitude but opposite in direction (since the person pushing the truck is not displaced when the truck doesn’t move). Therefore, the student’s logic is correct.
21. A hockey ball of mass 200 g travelling at 10 ms–1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 ms–1. Calculate the magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Solution:
Given, mass of the ball (m) = 200g
Initial velocity of the ball (u) = 10 m/s
Final velocity of the ball (v) = – 5m/s
Initial momentum of the ball = mu = 200g × 10 ms-1 = 2000 g.m.s-1
Final momentum of the ball = mv = 200g × –5 ms-1 = –1000 g.m.s-1
Therefore, the change in momentum (mv – mu) = –1000 g.m.s-1 – 2000 g.m.s-1 = –3000 g.m.s-1
This implies that the momentum of the ball reduces by 1000 g.m.s-1 after being struck by the hockey stick.