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A bullet of mass m moving with velocity v strikes a suspended wooden block of mass M.If the block rises to height h then the initialvelocity v of thebullet must have been what??

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3 years ago

```							The initial total energy (kinetic + potential) = the final total energy (kinetic + potential)     KE0 + PE0 = KEf + PEf     Right after 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) Now let's go back in time to look at the collision itself. For an inelastic collision (where the objects become stuck together), the initial momentum = final momentum (just like ANY other collision).      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 solved for earlier:     v0 = (m+M)*√(2gh)                   m I hope this helps!
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3 years ago
```							Answer Given that,Bullet of mass =mBlock of mass =MVelocity =vAs the bullet comes to rest with respect to the block, the two behaves as one body.Let V be the velocity of the combinationApplying the conservation linear momentum,  (m+M)V=mv+Mu V=(m+M)mv​.....(I)As block will rise to a height hPotential energy of combination = Kinetic energy of the combination  (m+M)gh=21​(m+M)V2 2gh=(m+Mmv​)2 v=mm+M​2gh​Hence, the initial velocity v is mm+M​2gh​
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5 months ago
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