Question

A 0.014 kg bullet traveling horizontally at 403 m/s strikes a 4 kg block of wood sitting at the very edge of a horizontal table of height 1 meters. The bullet is lodged into the wood. [For the signs of directions in this problem, take forward and upward to be positive; backward and down to be negative.]

a) What is the momentum of the bullet right before it impacts the block of wood?

b) What is the momentum of the bullet-and-block system immediately after the bullet is embedded within the block? ( (1612 kg·m/s, 1617.64 kg·m/s, 5.64 kg·m/s)

c) If the impact from the bullet knocks the bullet off of the edge of the table, what is the horizontal velocity of the bullet-and-block immediately after they start falling off the table? (403 m/s, 0 m/s. 1.41 m/s)

d) What is the vertical velocity of the bullet-and-block immediately after they start falling off the table?

Answer 1

Given the mass of the bulletz m= 0.014 kg

The velocity of the bullet, u = 403 m/s

a. The momentum of the bullet just before it hits the block, p = m*u

momentum ,p = 0.014* 403 = 5.64 kgm/s

So the momentum of the bullet = 5.64 kg m/s

b. Since the bullet is embedded in the block, it can be considered as a non elastic collision. And in a non elastic collision, the total momentum is conserved. That is total initial momentum before collision = total final momentum after collision.

Total initial momentum = momentum of bullet + momentum of block = 5.64+ 0

(0 since the block was at rest initially)

= 5.64 kgm/s

So the momentum of the system immediately after the bullet embedded = 5.64 kg m/s

C.

The Total mass of the system after collision,M = 4+0.014 = 4.014 kg

The momentum after collision =5.64 kg m/s

This momentum is the result of the horizontal velocity attained by the system by collision .

Let v be the horizontal velocity of the system

P = M* v

5.64 = 4.014 * v

v = 5.64/4.014 = 1.41 m/s

So the horizontal velocity of the system just after falling off = 1.41 m/s

D.

The collision has given only horizontal velocity to the system and not any vertical velocity since the bullet was travelling in horizontal direction at first.

So the vertical velocity of the system just after falling off the table = 0 m/s

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