An 10 g bullet travelling 310 m/s is shot into a 0.9 kg block of wood (initially

Question

An 10 g bullet travelling 310 m/s is shot into a 0.9 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 62 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) use conservation of momentum .01(310)=.01(62)+.8v v=3.1 m/s b)use conservation of energy (1/2)(.8)(3.1)2=(.8)(9.8)h h=.4903 m c) use sum of verticle forces T=9.8(.8) T=7.84 N d)now the block is in circular motion .8(3.1)2/2=T-.8(9.8) T=15.528 N Hope that helps

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