As shown in the figure below, a bullet is fired at and passes through a piece of

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 (0.486) after passing through the target. The collision is inelastic and during the collision, the amount of kinetic energy lost by the bullet and paper is equal to [(o.303)Kb Bcl, that is, 0.303 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= (a) Before collision (b) After collision

Explanation / Answer

momentum before collision momentum after collision

m*v = M*V + m*(0.486v)

V = (m/M)*0.514v

kinetic energy of bullet before collision

Kb Bc = (1/2)*m*v^2

kinetic energy of bullet after collision

KbAC = (1/2)*m*(0.486v)^2

kinetic energy of target after collision

ktAc = (1/2)*M*(V^2) = (1/2)*M*((M/m)*0.514v)^2

energy loss

KbBC - (Kt Ac + kbAc) = 0.303)*Kb BC

(1/2)*m*v^2 - ((1/2)*m*(0.486v)^2 + (1/2)*M*((m/M)*0.514v)^2) = (0.303)*(1/2)*m*v^2

1 - (0.486^2 + (m/M)*0.514^2) = (0.303)

M = 0.573 m

===========================

V = (m/M)*0.524*v

V = 0.914 v

part(a)

V = 0.914 v

part(b)

M = 0.573 m

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