A bullet (mass 0.0140 kg) hits a block of wood (mass 1.15 kg) and sticks in the

Question

A bullet (mass 0.0140 kg) hits a block of wood (mass 1.15 kg) and sticks in the wood. The wood is suspended by strings and proceeds to swing upward. It reaches a distance d above the point where it is struck by the bullet. Immediately after the collision and before any appreciable change in height, the bullet and block are moving at a speed of 2.25 m/s.

What is the speed of the bullet just before it collides with the wood?

What is the maximum height, d, reached by the bullet and block after the collision?

If the bullet travels into the block for a time of 7.20 ms before getting completely stuck, what is the average force that the bullet exerts on the block?

Suppose the bullet comes in at the same speed as above, but instead bounces backwards off the block. The masses are the same. Will the height, d, be: (a) larger than, (b) smaller than, or the (c) same as the initial experiment where the bullet is stuck in the wood?

Explanation / Answer

Common speed after collision, vc = 2.25 m/s

The bullet and the wood undergo inelastic collision.

Initial momentum = Final momentum

=> mbub = (mb + mw)vc

=> speed of bullet before collision, ub = [(mb + mw) / mb]vc = [(0.0140 + 1.15) / 1.15] * 2.25 = 2.28 m/s

The kinetic energy just after the collision is converted into the potential energy at height 'd'.

(mb + mw)vc2/2 = (mb + mw)gd

=> d = vc2/2g = 2.252 / (2 * 9.81) = 0.26 m

Change in momentum of bullet = Impulse

=> mbvc - mbub = Favgt

So, magnitude of average force applied by bullet on wood,

Favg = mb(ub - vc) / t = 1.15 * (2.28 - 2.25) / (7.2 * 10-3) = 4.8 N

If the bullet bounces backwards, then let's consider the momentum equation for this collision.

mbub = mwvw - mbvb

=> vw = mbub/mw + mbvb/mw > mbub/(mw + mb) + mbvb/mw = vc + mbvb/mw  

Hence, vw > vc

Hence, speed of wood after collision will be greater compared to before. So, the height wood reaches now will be larger than in the initial experiment.

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