As shown in the figure above, a bullet is fired at and passes through a paper ta

Question

As shown in the figure above, a bullet is fired at and passes through a paper target suspended by a massless string. The bullet has a mass m, a speed v before the collision with the target, and a speed (0.476)v after passing through the target. The collision is inelastic and during the collision, the amount of energy lost is equal to a fraction [(0.293)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.

v

v

Explanation / Answer

let the mass of target be M

Initial momentuj of bullet = mv

Initial momentum of target=0(because the target was at rest initially)

Total initial momentum of system = mv

Final momentum of bullet = 0.476mv

Final momentum of target=MV

Final momentum of system=0.476mv+MV

From conservation of momentum,

mv = 0.476mv+MV

Therefore,

MV = 0.524mv...(i)

initial kinetic energy of system =0.5*m*v2=0.5mv2

Final kinetic energy of systerm = 0.5m(0.476v)2+0.5MV2

Energy lost = 0.5mv2-0.226576mv2-0.5MV2=0.293*0.5mv2

Or, 0.126924mv2=0.5MV2

Or, MV2 = 0.2538mv2 ...(ii)

dividing equation (ii) by (i),we get

(MV2)/(MV) = (0.2538mv2)/ (0.524mv)

Or, V=0.484v

M = 1.08m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.