1.A 13 g bullet traveling 222 m/s penetrates a 2.0 kg block of wood and emerges

Question

1.A 13 g bullet traveling 222 m/s penetrates a 2.0 kg block of wood and emerges cleanly at 164 m/s. If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?
m/s

2. A 50 kg woman and an 85 kg man stand 11.0 m apart on frictionless ice.

(a) How far from the woman is their CM?
m

(b) If each holds one end of a rope, and the man pulls on the rope so that he moves 1.0 m, how far from the woman will he be now?
m

(c) How far will the man have moved when he collides with the woman?
m

Explanation / Answer

You can use conservation of linear momentum.

m1*u1 + m2*u2=m1*v1 + m2*v2 where m1 is the mass of the bullet, m2 is the mass of the wood, u1 and u2 are

the initial velocities of the bullet and the wood respectively, and v1 and v2 are the respective ending velocities

of the bullet and the wood.

Hence,

0.013 (convert g to kg)* 222 + 2*0 (the wood is at rest initially) = 0.013*164 + 2v2

2.886=2.132+2v2
0.754=2v2
v2=0.377 m/s

2)

a) by summing the moments around the woman, we find
(50kg+85kg)•x = 85kg•11m
x = 6.93 m

b) their c.o.g. does not move, so 1.1m/(11m - 6.93m) = x/6.93m
x = 1.87 m

c) She moves 6.93m; he moves 11m - 6.93m = 4.07m.

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