A 5.40 g bullet is fired horizontally at two blocks resting on a smooth tabletop

Question

A 5.40 g bullet is fired horizontally at two blocks resting on a smooth tabletop, as shown in the top figure. The bullet passes through the first block, with mass 1.20 kg, and embeds itself in the second, with mass 1.80 kg. Speeds of 0.340 m/s and 1.07 m/s, respectively, are thereby imparted to the blocks, as shown in the bottom figure. Neglecting the mass removed from the first block by the bullet, find the speed of the bullet immediately after it emerges from the first block. Then find the bullets original speed.

Explanation / Answer

Work backwards to find the speed of the bullet before (or as) it strikes and embeds itself into the second block. The law of conservation of momentum says that the initial momentum is equal to the final momentum. In equation form :

m?v?(i) + m?v?(i) = (m? + m?)v(f)

The second block is initially at rest, so its momentum is zero, and the equation becomes :

m?v?(i) = (m? + m?)v(f)
1)
Solved for v?(i) :

v?(i) = (m? + m?)v(f) / m?
= [(0.00540kg + 1.80kg)(1.07m/s)] / 0.00540kg
= 357.7 m/s

This is the speed of the block as it leaves block 1.
2)
For its speed as it enters block 1 :

m?v?(i) + m?v?(i) = m?v?(f) + m?v?(f)

In this case, m? is block 1, and its initial momentum is zero. The equation above, then, becomes :

m?v?(i) = m?v?(f) + m?v?(f)
v?(i) = [m?v?(f) + m?v?(f)] / m?
= [(0.00540kg)(357.7m/s) + (1.20kg)(0.340m/s)] / 0.00540kg
= 433.2m/s (b)

This is the speed of the bullet before it strikes block 1.

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