Physics
posted by Kerry on .
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.

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