The lead female character in the movie Diamonds Are Forever is standing at the e

Question

The lead female character in the movie Diamonds Are Forever is standing at the edge of an offshore oil rig. As she fires a gun, she is driven back over the edge and into the sea. Suppose the mass of a bullet is 0.0231 kg and its velocity is +693 m/s. Her mass (including the gun) is 53.7 kg. (a) What recoil velocity does she acquire in response to a single shot from a stationary position, assuming that no external force keeps her in place? (b) Under the same assumption, what would be her recoil velocity if, instead, she shoots a blank cartridge that ejects a mass of 0.000782 kg at a velocity of +693 m/s?

Explanation / Answer

Given:       The mass of the bullet, m = 0.0231 kg       The speed of the bullet, v = 693 m/s       The mass of bullet with female, M = 53.7 kg ------------------------------------------------------------------------------------ a)       The initial momentum of the femal is                                 Pi = 0       Using the law of conservation of momentum,                  0 = mv + MV       Therefore, the recoil velcity is                 V = -mv/M                     = -(0.0231 kg)(693 m/s)/(53.7 kg)                    = -0.2981 m/s                     = -0.30 m/s   (nearly) ------------------------------------------------------------------------------------ b)       The ejected mass, m = 0.000782 kg       The velocity, v = 693 m/s       Using the law of conservation of momentum,                  0 = mv + MV       Therefore, the recoil velcity is                 V = -mv/M                     = -(0.000782 kg )(693 m/s)/(53.7 kg)                    = -0.01009 m/s                     = -0.01 m/s   (nearly) ------------------------------------------------------------------------------------ b)       The ejected mass, m = 0.000782 kg       The velocity, v = 693 m/s       Using the law of conservation of momentum,                  0 = mv + MV       Therefore, the recoil velcity is                 V = -mv/M                     = -(0.000782 kg )(693 m/s)/(53.7 kg)                    = -0.01009 m/s                     = -0.01 m/s   (nearly)                  0 = mv + MV       Therefore, the recoil velcity is                 V = -mv/M                     = -(0.000782 kg )(693 m/s)/(53.7 kg)                    = -0.01009 m/s                     = -0.01 m/s   (nearly)

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