Question

Please answer both parts of question and show work.

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 (0.556)v after passing through the target. M (a) Before collision (b) After collision The collision is inelastic and during the collision, the amount of energy lost is equal to a fraction ((0.353)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 = M =

Answer 1

In elastic collision the momentum is con served According to the conservation of momentum, my= MV +m(0.556v) my-0.556mv= MV 0.444mv= MV The lost in kinetic energy is, [} mv +şm(0.556v)"}-{m* = 0.353KE mv = (0,691)m»* = 0,353 (} mv) MV2 - 1.044mv2 = 0 (0.444mv)V – 1.044mv = 0 0.444V -1.044v = 0 Hence velocity of the paper is, v_1.044 0.444 = 2.35V The mass of the paper is, 0.444mv= MV 0.444mv= M (2.35V) 0.444m = M(2.35) M-0.444m 2.35 = 0.187m = 0.187m