Question

Two balls collide at a 90-degree angle. The first ball has a mass of 1.5*10^4 kg and is moving with a velocity of 25m/s to the right. The second ball has a mass of 2.5*10^4 kg with a velocity of 35 m/s in a downward direction. They collide so that the first ball, the smaller one, hits the left side of the second ball and continues to move forward with a fifth of its original speed. What is the new speed and direction of the larger asteroid?

Answer 1

Solution:

Momentum before collision = Momentum after collision according to momentum conservation

X direction velocity  of m1 1.5 x 10^4 kg = v1x = 25 m/s

Y direction velocity of m1 = 0= v1y

X direction velocity of m2 2.5 x 10^4 kg = v2x = 0

Y direction velocity of m2 = v2y = -35 m/s

X direction momentum conservation :

1.5x10^4 (25) + 2.5x10^4 *0 = 1.5x10^4 (25/5) +(2.5×10^4)Vx

=> Vx = 1.5x10^4 (25-5) /(2.5×10^4) = 12 m/s

Y direction momentum conservation :

1.5x10^4 *0 + 2.5x10^4 (-35) = 0 + 2.5x10^4 * Vy

=> Vy = -2.5x10^4 × (-35) / 2.5x10^4 = -35 m/s

total velocity of the heavier ball = Vx^2 + Vy^2 = 12^2 + (-35)^2

= 37 m/s

direction = tan^-1( -35 /12) = 71 degrees, clockwise from +x axis