Question

2.47 is not the correct answer

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 (0.506)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 [(0.313)KE, Bc) of the kinetic energy of the bullet before the collision. Determine the mass M of the target and the speed 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 2.47 X = What physical quantities are conserved as the bullet hits and passes through the target? Represent the mass and speed of the target after collision by Mand v and then see if you can write an equation which expresses the momentum of the target after the collision in terms of the mass and speed of the incident bullet. After determining the amount of kinetic energy lost, see if you can write an equation which expresses the kinetic energy of the target after the collision in terms of the mass and speed of the incident bullet. At this point you have two equations and the mass and speed of the target after the collision as unknowns. See if you can solve these two equations simultaneously to obtain the mass and speed of the target in terms of the mass and initial speed of the bullet.v M0.2008 x = What physical quantities are conserved as the bullet hits and passes through the target? Represent the mass and speed of the target after collision by Mand V and then see i you can write an equation which expresses the momentum of the target after the collision in terms of the mass and speed of the incident bullet. After determining the amount of kinetic energy lost, see if you can write an equation which expresses the kinetic energy of the target after the collision in terms of the mass and speed of the incident bullet. At this point you have two equations and the mass and speed of the target after the collision as unknowns. See if you can solve these two equations simultaneously to obtain the mass and speed of the target in terms of the mass and initial speed of the bullet.m

Answer 1

conservation momentum, Apply mis-mutmine mv = mv (1-0.506) = 0.494mve Kf-ka=0.313 kb ka 1-0.313 Kb= 0.687. Kb. & MUZ5MY (0.687-(0-506)) my²x 0.430 Hv= myx V 0.494 h = myx0.430 V V=10.87 v M = 0.494 m = 0.567 0.567 m 0.87