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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??

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??    

Grade:11

2 Answers

Arun
25750 Points
6 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!
ankit singh
askIITians Faculty 614 Points
3 years ago

Answer

 

Given that,
Bullet of mass =m
Block of mass =M
Velocity =v
As the bullet comes to rest with respect to the block, the two behaves as one body.
Let V be the velocity of the combination
Applying the conservation linear momentum,
  (m+M)V=mv+Mu
 V=(m+M)mv​.....(I)
As block will rise to a height h
Potential 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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