A block of mass M hangs vertically from light strings. The tops of the string ar
ID: 2270241 • Letter: A
Question
A block of mass M hangs vertically from light strings. The tops of the string are firxed in a laboratory near the surface of the earth. Initially at stage 1, the block is at rest. A bullet of mass m and inital horizontal velocity v1, strikes the block. The bullet becomes embedded in the block, and causes the block to move to the right, at stage 2. The block and bullet combination then swings to the right and upward. It comes momentarily to rest at a vertical distance h up from its original location, stage 3.
a) consider the collision between the bullet and the block (transition between stage 1 and stage 2) Assume that the collision is instantaneous. Use conservation of momentum to find the velocity of the ball and block combination immediately after the collision. What is v2 in terms of m, M, and v1?
b) Consider the swinging process in which the bullet and block move from stage 2 to stage 3. Use conservation of energy to find the height, h, to which the system rises before coming to rest? Write your answer in terms of m, M, v2, and g.
c) Substitute your answer for part a into you answer for part b and find height, h, in terms of m, M, v1, and g. The answer should be: h= 1/2g (mv1/m+M)^2
Please explain everything thoroughly. Thank you.
of the initially ass m ecomes ht ght and p from sition city of In move height answerExplanation / Answer
a) using conservation of momentum,we have
momentum of bullet before collision=momentum of whole system after collision.i.e.
mv1=(m+M)v2
hence v2=mv1/(m+M)
b)using conservation of energy we have
energy of whole system just after collision(which is kinetic energy)=energy when system comes to instantaneous rest(which is potential energy).i.e.
initial kinetic energy=final potential energy
1/2*(m+M)*v2^2=(m+M)*g*h
hence h=v2^2/2g.
c)putting v2=mv1/(m+M)
h={mv1/(m+M)}^2/2g
=1/2g*(mv1/m+M)^2.
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