A bullet of mass m = 0.050 kg with a speed v = 50 m/s hits a wooden target of ma

Question

A bullet of mass m = 0.050 kg with a speed v = 50 m/s hits a wooden target of mass M = 0.50 kg. The target is suspended by a string of negligible mass and of length 2.0 m. After the collision between the bullet and the target, the bullet penetrates the target and comes out with a velocity v' = 10 m/s following a straight path, and the target follows a circular path as shown in Figure. What is the velocity of the target V right after the collision? How high does the target reach from the initial height?

Explanation / Answer

a) using conservation of momentum,

mv + Mu =mv' + MV (u=0)

0.05*50 + 0.5*0 = 0.05*10 + 0.5 * V

V= 4 m/s

b) Now 0.5 m V2 =mgh

h=0.5*16/9.8 =0.816 m

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