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 (0.496)v after passing through the target. The collision is inelastic and during the collision, the amount of energy lost is equal to a fraction [(0.423)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.)
http://www.webassign.net/webassignalgphys1/8-p-035.gif 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 (0.496)v after passing through the target. The collision is inelastic and during the collision, the amount of energy lost is equal to a fraction [(0.423)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.)
http://www.webassign.net/webassignalgphys1/8-p-035.gif 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 (0.496)v after passing through the target. The collision is inelastic and during the collision, the amount of energy lost is equal to a fraction [(0.423)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.)

Explanation / Answer

conservation of momentum

m v = m*0.496v + M V

M = 0.504 mv/V

KE loss = KE final - KE initial

-0.423* 1/2 mv^2 = ( 1/2 M V^2 +0.5* m*(0.496 v)^2 ) - 1/2 m v^2

plug in M

-0.423 m v^2 = 0.504 m v V + 0.496^2 m v^2 - m v^2

m cancels

-0.423 v^2 - 0.496^2 v^2 + v^2 = 0.504 v V

V = ( -0.423 - 0.496^2 +1)/0.504 v=0.657 v

so M = 0.504 m v/(0.657 v) = 0.504/0.657 m=0.767 m

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