Question

A large crate is suspended by a light string. A bullet is fired horizontally into the crate and becomes firmly lodged inside it. After being struck by the bullet, the crate swings upward to a maximm height and then swings back up.

Just before the collision (time t1), the crate is at rest and the bullet moves horizontally with speed v0. Immediately after the bullet becomes lodged inside the crate (time t2) the bullet and crate move together with speed v. The crate reaches its maximum height at time t3.

1. Consider the time interval between times t1 and t2 (i.e., the collision between the bullet and crate).

A) During the collision, how does the force exerted on the crate by the bullet compare to the force exerted on the bullet by the crate? Discuss both magnitude and direction. Explain?

B) Is the total momentum of the bullet-and-crate system conserved during the collision? Explain how you can tell.

C) Is the total kinetic energy of the system conserved during the collision? Explain how you can tell.

2. Now consider the time interval from t2 to time t3 for the situation described (when the bullet and crate move together to a maximum height after the collision).

A) Draw a free body diagram for the bullet-crate system for an instant between times t2 and t3. Clearly label all forces.

Is the total momentum of the system conserved from tme t2 to time t3? Explain.

B) For each force you indicated on your free body diagram in part a, indicate whether the work done by that force on the bullet-crate system is positive, negative, or zero. Explain your reasoning.

Is the total mechanical energy (kinetic + potential) of the bullet-crat system conserved from time t2 to time t3? Explain.

3. Is it incorrect to say that the total mechanical energy of the bullet-crate system remains conserved for the entire motion. Explain why this is so, and indicate other types of energy into which the initial energy of the bullet could have been transformed.

Answer 1

1)

A)

Using newton's third law , force exerted by crate on bullet must be equal in magnitude to the force exerted by the bullet on crate and direction is exactly opposite.

B)

after the collision, the bullet and crate moves together. Along the direction of motion , their is no net external force , hence since the collision is elastic and in elastic collision , we have momentum conserved

C)

since the collision is inelastic , Kinetic energy will not be conserved. some energy is transformed in frictional losses and heat

2.

A)

M = mass of crate

m = mass of bullet

T = tension in rope

B)

Tension force acts in perpendicular direction to the direction of motion of crate-bullet at any instant of time.

hence work done by tension force = 0

gravity force acts down but the displacement is upwards , hence work done by gravity force or weight is negative

Total mechanical energy of bullet-crate system is conserved . the kinetic energy at the bottom just after the collision converts completly to potenial energy it gains at the topmost point

3.

yes it is correct since during the collision some energy is lost to heat and friction and hence kinetic energy is not conserved there.