Posted by **Kerry** on Tuesday, February 12, 2013 at 3:02pm.

A 0,1kg bullet is fired horizontally towards blcok A and B, which are hanging from cords. The bullet passes through block A of mass 2kg and becomes embedded in block B of mass 6.4kg. The bullet thus causes A and B to start moving with velocities of 6m,s- and 2m.s- respectively, imeediately after impact. Calculate:

a) the initial velocity of the bullet

b) the velocity of the bullet as it travels from A to B.

- Physics -
**Ajit**, Tuesday, March 19, 2013 at 12:05am
Assume u = Initial velocity of the bullet and v= its vel after leaving block A.

V = Vel. of block A after the impact = 2 met/sec

M = Mass of block A = 2 kg & m = mass of the bullet = 0.1 kg

By the principle of conservation of momentum: m*u = m*v + M*V

Or 0.1u = 0.1v + 2*6

Applying the same principle to block B: 0.1v = (6.4 +0.1*2

which gives us: v =130 met/sec

Plug this value in the earlier equation to obtain: u = 250 met/sec

Thus, initial bullet vel. =250 met/sec and Buleet vel. as travels from A to B = 130 met/sec

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