An 7 g bullet travelling 330 m/s is shot into a 0.8 kg block of wood (initially

Question

An 7 g bullet travelling 330 m/s is shot into a 0.8 kg block of wood (initially at rest) suspended from the ceiling by a 2 m long light rope. The bullet exits the block at a velocity of 66 m/s.
(a) What is the velocity of the block of wood just after the bullet exits the block?
m/s

(b) How high (above the starting position) does the block reach after the collision?
m

(c) Immediately before the collision, what is the tension in the rope?
N

(d) Immediately after the collision, what is the tension in the rope?
N

Explanation / Answer

a)0.5 * 0.8 * v^2 = 0.5 * 0.007 * (330^2 - 66^2) so V = 30.26 m/s b) let the block reaches a height h mgh = 0.5*m*v^2 h = 45.78 m c) T = mg = 0.8*10 = 8 N d) just after collision T - mg = mV^2/r T = 374.26 N

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