At a wildwest show a marksman fires a bullet at a 12 g coin

At a wild-west show, a marksman fires a bullet at a 12 g coin that\'s thrown straight up into the air. The marksman points his rifle at a 45 angle above the ground, then fires a 15 g bullet at a speed of 580 m/s . Just as the coin reaches its highest point, the bullet hits it and glances off, giving the coin an exactly vertical upward velocity of 150 m/s

At what angle measured with respect to the horizontal does the bullet ricochet away from this collision?

Solution

At the highest point, veolcity of the bullet is

Vbullet = V * cos(theta) i   = 580cos(45) i

The coin goes in perpendicular direction at Vcoin = 150 j m/s

Therefore conserving momentum in vertical direction

Mcoin * Vcoin = -Mbullet * Vbullet_y

Vbullet_y = -12 * 150 / 15   = -120 m/s

Tan(angle) = Vy / Vx   = -120 / 580*cos(45)  

angle = -16.30926 degrees with the horizontal