A bullet of mass m_b = 3.5 times 10^3 kg is fired horizontally at two blocks Mi

Question

A bullet of mass m_b = 3.5 times 10^3 kg is fired horizontally at two blocks Mi = 1.2 kg and M_2 = 1.8 kg. The two blocks are sitting on a horizontal smooth surface. The bullet passes through M_1 and embeds itself on M_2. This results in Mi moving to the right with a speed of 0.63 m/s and M_2 plus the bullet moving to the right with a speed of 1.4 m/s as shown in the diagram. What is the initial velocity of the bullet before encountering M_1 and M_2? What is the velocity of the bullet immediately after it emerges from the first block (M_1)?

Explanation / Answer

The collision with block 2 is totally inelastice (so Kinetic Energy is not conserved).
mbullet*vbullet = (mbullet+m2)*v2
vbullet = (.0035 + 1.8)*1.4/.0035 = 721.4/s

So the bullet had a speed of 721.4m/s just before it hit block number 2, which is the speed it had when it left block number 1.

B) mbullet*vbullet = mbullet*721.4+ m1*v1
vbullet = (0.0035*721.4 +1.2*.630)/.0035 = 937.4m/s

So the bullet was travelling at 937.4/s before it hit block #1.

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