Speed after a collision...?

Question

A 80 kg football player is stationary when a 100 kg linebacker hits him moving at 8 m/sec. After the collision, the linebacker is moving in the same direction at 1 m/sec. What is the speed of the first player after the collision? What percent of the energy is lost during the collision? Can you explain your answer?

Answer 1

This is a conservation of momentum problem. The total momentum before and after the collision is the same. I assume the football players don't bounce off each other, so the collision is inelastic (they stick together), and kinetic energy is not conserved.

Total momentum is m1*v1 + m2*v2, where m1, m2 are the footballer's masses and v1, v2 are their respective velocities.

Before collision, v2=0 so total mometum is m1^v1
After collision, both players are moving as one at rate v3, and their combined momentum is (m1+m2)*v3; conserving momentum means

m1*v1 = (m1+m2)*v3

You know m1, m2 and v1, solve for v3

The total kinetic energy before collision is .5*m1^v1^2; after collision it is .5*(m1+m2)*v3^2. Subtract to see how much energy is lost.

Answer 2

Momentum before collision = 100 * 80 = 800.
Momentum of line backer after collision = 100*1 = 100
Momentum of the player after collision = 800 - 100 = 700. Velocity = 700/80 = 8.75 m/sec

Kinetic energy before collision = 0.5*100*8^2 = 3200.
KE after collision = 0.5 * 100 * 1^2 + 0.5 * 80 * 8.75^2 = 3112.5
Lost energy = 3200 - 3112.5 = 87.5 J

Answer 3

your question is not complete. you have to specify what kind of a collision it is (elastic or inelastic).
if it is an elastic collision, no energy would be lost. else, it would depend on the COEFFICIENT of RESTITUTION.
For the first part of the question, if it were an elastic collision,
By conservation of momentum,
M1U1+M2U2 = M1V1+M2V2
80x0+100x8 = 80xV1+100x1
so, V1 = 700/80= 8.75m/s in the direction of the line backer (since the answer is +ve).
Note: this is true only for a head-on collision

Answer 4

conservation of momentum
(mL)(uL) x (mP)(uP) = (mL)(vL) x (mP)(vP)
(100)(8) x (80)(0) = (100)(1) x (80)(vP)
(vP) = 8.75m/s

initial KE - final KE = energy loss
1/2 (mL)(uL)^2 - (1/2 (mP)(vP)^2 + 1/2 (mL)(vL)^2)
= 1/2 (100)(8)^2 - (1/2(80)(8.75)^2 + 1/2(100)(1)^2 )
=3200- (3062.5 + 50)
=87.5 J

87.5/ 3200 X 100% = 2.73% energy loss

yey =)