a bullet of mass m moving with velocity v strikes a suspended wooden block of mass M if the block rises to a height h the initial velocity of the block will be
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Solution
The initial total energy (kinetic + potential) = the final total energy (kinetic + potential)
KE0 + PE0 = KEf + PEf
the bullet hits the block, the initial height is 0, so PE0=0. When the bullet/block reaches its maximum height, it is at rest, so KEf=0.
KE0 = PEf
(1/2)(m+M)vf2 = (m+M)gh
We can use this equation to solve for "vf", the velocity of the bullet/block after the collision:
vf = √(2gh)
. For an inelastic collision (where the objects become stuck together), the initial momentum = final momentum
p0 = pf
mbullet*v0,bullet + Mblock*v0,block = (mbullet + Mblock)*vf
Since the block is not moving before the collision, v0,block = 0
mbullet*v0,bullet = (m+M)*vf
Rearranging to solve for in initial velocity of the bullet, "v0":
v0 = (m+M)*vf
m
Substituting in the expression for "vf" that we
v0 = (m+M)*√(2gh)
m
The initial total energy (kinetic + potential) = the final total energy (kinetic + potential)
KE0 + PE0 = KEf + PEf
the bullet hits the block, the initial height is 0, so PE0=0. When the bullet/block reaches its maximum height, it is at rest, so KEf=0.
KE0 = PEf
(1/2)(m+M)vf2 = (m+M)gh
We can use this equation to solve for "vf", the velocity of the bullet/block after the collision:
vf = √(2gh)
. For an inelastic collision (where the objects become stuck together), the initial momentum = final momentum
p0 = pf
mbullet*v0,bullet + Mblock*v0,block = (mbullet + Mblock)*vf
Since the block is not moving before the collision, v0,block = 0
mbullet*v0,bullet = (m+M)*vf
Rearranging to solve for in initial velocity of the bullet, "v0":
v0 = (m+M)*vf
m
Substituting in the expression for "vf" that we
v0 = (m+M)*√(2gh)
m
malkitsinghbatth45:
Thank you so much
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