Engineering Physics Questions and Answers Part-13

1. A machine gun fires 60 bullets per minute, with a velocity of 700m/s. If each bullet has a mass of 50g, find the power developed by the gun.
a) 1225W
b) 12250W
c) 122.5W
d) 122W

Answer: b
Explanation: Mass of the bullets = 60×50 = 3000g = 3kg
v = 700m/s t = 1min = 60s
Power = W/t = (Kinetic energy)/t = 12250W.

2. For a collision to occur, the actual physical contact is necessary.
a) True
b) False

Answer: b
Explanation: A collision is said to occur between two bodies, either if they physically collide against each other or if the path of one is affected by the force exerted by the other. For a collision to take place, the actual physical contact is not necessary.

3. Which of the following is an example of inelastic collision?
a) Collision between two vehicles
b) Collision between glass balls
c) A bullet fired into a wooden block
d) Collision between two railway compartments

Answer: a
Explanation: If there is a loss of kinetic energy during a collision, it is called an inelastic collision. Collision between two vehicles is an example for inelastic collision.

4. Mud thrown on a wall and sticking to it is an example for ___________
a) Inelastic collision
b) Elastic collision
c) Super elastic collision
d) Perfectly inelastic collision

Answer: d
Explanation: If two bodies stick together after the collision and move as a single body with a common velocity, then the collision is said to be perfectly inelastic collision. A mud thrown on a wall sticks to the wall, hence it is an example for perfectly inelastic collision.

5. Collision between two carom coins is an example for ___________
a) Oblique collision
b) Perfectly inelastic collision
c) Inelastic collision
d) Elastic collision

Answer: a
Explanation: If two bodies do not move along the same straight line path but lie in the same plane before and after the collision, the collision is said to be oblique or two dimensional collisions.

6. When a light body collides with a massive body at rest ___________
a) The light body rebounds after the collision
b) The light body Remains at rest
c) The massive body rebounds after the collision
d) No reaction happens

Answer: a
Explanation: When a light body collides against a massive body at rest, the light body rebounds after the collision with an equal and opposite velocity while the massive body practically remains at rest. A light ball striking a wall rebounds almost with the same speed and the wall remains at rest.

7. When a massive body collides against a light body at rest ___________
a) The light body starts moving
b) The light body rebounds
c) The velocity of the bodies get exchanged
d) The massive body comes to rest

Answer: a
Explanation: When a massive body collides against a light body at rest, the velocity of the massive body remains almost unchanged while the light body starts moving with a velocity wise the velocity of the massive body.

8. If a force acts perpendicular to the direction of motion of a body, what is the amount of work done?
a) Infinity
b) Constant
c) Zero
d) Sinθ

Answer: c
Explanation: If a force acts perpendicular to the direction of a body, the amount of work done is zero because there is no displacement in the direction of a force.

9. A body can have momentum without energy.
a) True
b) False

Answer: b
Explanation: If a body has momentum, it must be in motion and consequently possess kinetic energy. Therefore a body cannot have momentum without energy.

10. Comets mover around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every orbit of the comet is zero. Why?
a) The gravitational force is conservative, hence work done is zero
b) The gravitational force is non-conservative, hence work done is zero
c) Energy is absent, hence work done is zero
d) Force is in negative direction

Answer: a
Explanation: The gravitational force acting on the comet is a conservative force. The work done by a conservative force is equal to the negative of the change in potential energy. Over a complete orbit of any shape, there is no change in potential energy of the comet. Hence no work is done by the gravitational force on the comet.