As shown in the figure below, a bullet is fired at and passes through a piece of
ID: 2003279 • Letter: A
Question
As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. The bullet has a mass m, a speed v before the collision with the target, and a speed (0.456)v after passing through the target. The collision is inelastic and during the collision, the amount of energy lost is equal to a fraction [(0.383)KEb BC] of the kinetic energy of the bullet before the collision. Determine the mass M of the target and the speed V of the target the instant after the collision in terms of the mass m of the bullet and speed v of the bullet before the collision. (Express your answers to at least 3 decimals.)V = v M = m
As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. The bullet has a mass m, a speed v before the collision with the target, and a speed (0.456)v after passing through the target. The collision is inelastic and during the collision, the amount of energy lost is equal to a fraction [(0.383)KEb BC] of the kinetic energy of the bullet before the collision. Determine the mass M of the target and the speed V of the target the instant after the collision in terms of the mass m of the bullet and speed v of the bullet before the collision. (Express your answers to at least 3 decimals.)
V = v M = m
(0.456)v [(0.383)KEb BC] V = v M = m
Explanation / Answer
Givn Vb=0.456v
By Conservation of momentum
mv = m*Vb+MV
mv =m*0.456v +MV
=>MV =(1-0.456)mv
MV = 0.544mv
M=(0.544mv/V)--------1
By Conservation of energy
KEbefore =KEafter+Energy lost
(1/2)mv2 = (1/2)mVb2+(1/2)MV2+0.383KEb (Before collision)
(1/2)mv2 =(1/2)m(0.456v)2 + (1/2)MV2+0.383*(1/2)mv2
mv2[1-0.208-0.383] =MV2
0.409mv2 =MV2
0.409mv2=(0.544mv/V)V2
=>V=0.752v
From 1
=>M=(0.544mv/0.752v)
M=0.724m
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