Question 7: A 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;
(b) the acceleration of the train; and
(c) the force of wagon 1 on wagon 2.
Answer: Given, force of engine = 40000 N
Force of friction = 5000 N
Mass of engine = 8000 kg
Total weight of wagons = 5 x 2000 kg = 10000 kg
(a) The net accelerating force
= Force exerted by engine – Force of fricition
= 40000 N – 5000 N = 35000 N
(b) The acceleration of the train
We know that, F = mass x acceleration
Or, 35000 N = (mass of engine + mass of 5 wagons) X a
Or, `35000 N = (8000 kg + 10000 kg) xx a`
Or, `35000N = 18000 kg xx a`
`=> a=(35000N)/(18000\ kg)=1.944ms^(-2)`
(c) The force of wagon 1 on wagon 2 = mass of four wagons x acceleration
`=> F= 4xx2000\ kg xx 1.944ms^(-2)`
`=>F= 8000\ kg xx 1.944ms^(-2)`
Thus, (a) The net accelerating force = 35000N
(b) The acceleration of train = 1.944 ms–2
(c) The force of wagon 1 on 2 = 1552 N
Question 8: 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 m s-2?
Answer: Given, Mass of the vehicle, m = 1500 kg
Acceleration, a = – 1.7 m s–2
Force acting between the vehicle and road, F =?
We know that, `F = m xx a`
Therefore, `F = 1500 kg xx 1.7 m s^-2`
Or, `F =-2550 N`
Thus, force between vehicle and road = - 2550 N. Negative sign shows that force is acting in the opposite direction of the vehicle.
Question 9: What is the momentum of an object of mass m, moving with a velocity v?
(a) (mv)2 (b) mv2 ( c ) Ω mv2 (d) mv
Answer: (d) mv
Explanation: Given, mass = m, velocity = v, therefore, momentum =?
We know that, momentum, P = mass x velocity
Therefore, P = mv
Thus, option (d) mv is correct
Question 10: 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?
Answer: Since, a horizontal force of 200N is used to move a wooden cabinet, thus a friction force of 200N will be exerted on the cabinet. Because according to third law of motion, an equal magnitude of force will be applied in the opposite direction.
Question 11: 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 m s-1 before the collision during which they stick together. What will be the velocity of the combined object after collision?
Answer: Since, two objects of equal mass are moving in opposite direction with equal velocity, therefore, the velocity of the objects after collision during which they stick together will be zero.
Explanation: Given, Mass of first object, m1 = 1.5 kg
Mass of second object, m2 = 1.5 kg
Initial velocity of one object, u1 = 2.5 m/s
Initial velocity of second object, u2 = -2.5 m/s (Since second object is moving in opposite direction)
Final velocity of both the objects, which stick after collision, v =?
We know that,
`=>1.5\ kg xx 2.5ms^(-1)` `+1.5\ kg xx(-2.5ms^(-1))` `=1.5\ kg xx v + 1.5\ kg xx v`
`=>3.75\ kg ms^(-1)-3.75\ kg\ ms^(-1)` `=v(1.5\ kg+1.5\ kg)`
Therefore, final velocity of both the objects after collision will be zero.
Question 12: 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.
Answer: Because of the huge mass of the truck, the force of static friction is very high. The force applied by the student is unable to overcome the static friction and hence he is unable to move the truck. In this case, the net unbalanced force in either direction is zero which is the reason of no motion happening here. The force applied by the student and the force because of static friction are cancelling out each other. Hence, the rationale given by the student is correct.
Question 13: A hockey ball of mass 200 g travelling at 10 m/s is struck by a hockey stick so as to return it along its original path with a velocity at 5 m/s. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Answer: Given, Mass of hockey ball, m = 200 g = 200/1000 kg = 0.2 kg
Initial velocity of hockey ball, u = 10 m/s
Final velocity of hockey ball, v = – 5 m/s (because direction becomes opposite)
Change in momentum =?
We know that,
Momentum = mass x velocity
Change in momentum `= m(v - u)`
`= 0.2 kg(-5 m//s - 10 m//s)`
`= 0.2 xx(-15)= -3 kg m//s`
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