Question

IP A charging bull elephant with a mass of 5100 kg comes directly toward you with a speed of 4.50 m/s . You toss a 0.200-kg rubber ball at the elephant with a speed of 7.60 m/s .

What is the speed when it bounces back at you

Answer 1

A charging bull elephant with a mass of 5100 kg comes directly toward you with a speed of 4.50 m/s . You toss a 0.200-kg rubber ball at the elephant with a speed of 7.60 m/s .

What is the speed when it bounces back at you

Assume the elephant is running from right to left (the minus direction), and the ball therefore is thrown in the positive direction. Direction is important in collision problems. This is an elastic collision, in this type of collision, both momentum and kinetic energy is conserved. Call the mass of the ball m₁ and the elephant’s mass m₂. From the law of conservation of momentum:

m₁v₁(i) + m₂v₂(i) = m₁v₁(f) + m₂v₂(f)
(0.200kg)(7.60m/s) + (5,100kg)(-4.50m/s) = (0.200kg)v₁(f) + (5,100kg)v₂(f)
-22,948.48 kg∙m/s = (0.200 kg)v₁(f) + (5,100 kg)v₂(f)--------------------->(1)

In order to find the ball’s final velocity, v₁(f), we need another equation. The laws of conservation of energy and momentum may be combined to give an equation that relates the velocities of approach to the velocities of recession of two colliding bodies. The derivation of the equation will likely be in your textbook. The equation is:

v₁(i) - v₂(i) = v₂(f) - v₁(i)
7.60m/s -(-4.50m/s) = v₂(f) - v₁(i)
12.1m/s = v₂(f) - v₁(i)
v₂(f) = v₁(f) + 12.1 m/s------------------->(2)

Plugging (2) into (1) eliminates v₂(f) and allows you to find v₁(f):

-22,948.48 kg∙m/s = (0.200 kg)v₁(f) + (5,100 kg)[ v₁(f) + 12.1 m/s]
-22,948.48 kg∙m/s = (0.200 kg)v₁(f) + (5,100 kg)v₁(f) + 61,710 kg∙m/s
(5,100.2 kg)v₁(f) = -84658.48 kg∙m/s
v₁(f) = -16.599 m/s

So the ball's velocity is 16.599 m/s in the opposite direction it was initially thrown.