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A bullet of mass 0.065 kg traveling horizontally at a speed of 250 m/s embeds itself in a block of mass 1.5 kg that is sitting at rest on a nearly frictionless surface. What is the speed of the block after the bullet embeds itself in the block?
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See the second bullet point here.
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For this question, you can solve it simply using the conservation of momentum.
Before Collision:
Bullet:
mass= 0.065 kg
v= 250m/s
Block:
mass=1.5 kg
v= 0
After Collision:
Bullet and Block sticks together:
mass= 1.565 kg
v= ???
sum of momentum before= sum of momentum after
(0.065 kg)(250 m/s)+(1.5 kg)(0)= (1.565 kg)(v)
16.25 kg*m/s=1.565 kg*v
v= 10.38 m/s
But next time when you ask a question, at least show your work so that others know you've put in an effort to try it.
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Nice work seigakufuji,I wouldn't have put the sticking of the bullet and block into consideration
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Woo is this For Real guys?
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