Physics help?

Question

A 0.170 kg projectile is fired with a velocity of +710 m/s at a 2.00 kg wooden block that rests on a frictionless table. The velocity of the block, immediately after the projectile passes through it, is +45.0 m/s. Find the velocity with which the projectile exits from the block.

Can someone help me with this problem?

Evil Monkey rocks!

Answer 1

calculate the momentum of the projectile

p=mv
p=(0.17 kg)(710 m/s)
p=120.7 kgm/s

calculate the momentum of the wooden block

p=mv
p=(2 kg)(45 m/s)
p=90 kgm/s

since momentum cannot be destroyed or created in a closed theoretical system:

120.7 kgm/s - 90 kgm/s = 30.7 kgm/s

the initial momentum is 120.7 but once the projectile makes contact with the wooden block, it transfers 90 of its original momentum to the block...leaving 30.7 left for the projectile

to solve the velocity:

p=mv
30.7=(0.17 kg) v
v=180.6 m/s

Answer 2

Great analysis from Evil Monkey. He's not evil after all.

It's a conservation of momentum problem. Just equate the initial momentum of the projectile, to the momentum of the block and the momentum of the projectile after it passes through the block. You should get the same answer.

Answer 3

m1v1+m2v2=mfvf

Masses and velocities, just plug and chug.