Question

A 77.8 kg basketball player is running in the positive direction at 8.1 m/s. She is met head-on by a 99.8 kg player traveling at 6.9 m/s toward her. If the 99.8 kg player is knocked backwards at 3.5 m/s, what is the resulting velocity of the 77.8 kg player?

Answer 1

Answer:

-5.24 m/s

** The minus sign indicates that the velocity vector points in the opposite direction with respect to the initial direction of the 77.8 kg player **

Explanation:

Hi!

We can solve this problem considering each player as a point particle and taking into account the conservation of linear momentum.

Since the 99.8 kg player is moving towards the 77.8kg, the initial total momentum is:

m1*v1_i + m2*v2_i = (77.8kg)(8.1 m/s) - (99.8kg)(6.9 m/s)

** The minus sign indicates that the velocity vector points in the opposite direction with respect to the initial direction of the 77.8 kg player **

The final total momentum is equal to:

m1*v1_f + m2*v2_f = (77.8 kg)v1_f + (99.8 kg)(3.5 m/s)

The conservation of momentu tell us that:

m1v1_i + m2v2_i = m1v1_f + m2v2_f

Therefore:

v1_f =v1_i + (m2/m1)*(v2_i-v2_f)

v1_f = 8.1 m/s  + (99.8 / 77.8) * (-6.9 - 3.5 m/s)

v1_f = -5.24 m/s