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 (0.516)v after passing through the target. M (a) Before collision M (b) After collision The collision is inelastic and during the collision, the amount of energy lost is equal to a fraction ((0.283)KE 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

Answer 1

conservation of momentum in horizontal direction, m (0.5161) + MV -0 mo = Also I I mor- ( 1 m (ostav?? + MV 2 ) 2 ( ) 1 muz © 0.283 ( fraction of KE lost) From O MU - 0.484 mv From @ I - 0.266 M v² 0.283 m 62 = Mv2 0.451 - mo = 0-45 1 x mor 0.45lXmv2 MV 0.484 mo Va -0.932u IM = 0.52m