A projectile with a mass 6M is fired at a speed of 400 m/s at an angle of 60^o above the horizontal. At the highest point of its trajectory, the projectile is broken into two equal pieces by an internal explosion. Just after the explosion, one of the two pieces is known to be traveling vertically downward at a speed of 300 m/s. The magnitude of the velocity of the other half of the projectile is A) 500 m/s. B) 1.50 km/s. C) 400 m/s. D) 710 m/s. E) 123 m/s.

Question

Quizplus · Accepted Answer

Explanation:Momentum is conserved in the explosion. The total momentum before the explosion is:$$P_i = 6M(400 m/s) \hat{i} + 6M(400 m/s) \sin 60^o \hat{j}$$The momentum after the explosion is:$$P_f = 3M(300 m/s) \hat{j} + M(v) \hat{i} + M(v) \hat{j}$$Setting these two expressions equal to each other, we get:$$6M(400 m/s) \hat{i} + 6M(400 m/s) \sin 60^o \hat{j} = 3M(300 m/s) \hat{j} + M(v) \hat{i} + M(v) \hat{j}$$Solving for v, we get:$$v = 500 m/s$$

A Projectile with a Mass 6M Is Fired at a Speed