2. On a touchdown attempt, 95.00 kg running back runs toward the end zone at 3.7

Question

2. On a touchdown attempt, 95.00 kg running back runs toward the end zone at 3.750 m/s. A 113.0 kg line-backer moving at 5.380 m/s meets the runner in a head-on collision. If the two players stick together, a) what is their velocity immediately after collision? b) What is the kinetic energy of the system just before the collision and a moment after the collision? These are the steps, I just do not understand what they are doing. Please show work with the actual numbers from the problem. Thanks. miVimzVi negative sign indicates that final velocity vector is in same direction as the initial velocity of line-backer

Explanation / Answer

Part A.

By using momentum conservation we got that

Vf = (m1v1 - m2v2)/(m1 + m2)

m1 = mass of running back = 95 kg

m2 = mass of line backer = 113 kg

v1 = initial speed of running back = 3.750 m/sec

v2 = initial speed of line backer = -5.380 m/sec (-ve sign because both are running in opposite direction)

So,

Vf = (95*3.750 - 113*5.380)/(95 + 113) = -1.210 m/sec

So after collision velocity will be 1.210 m/sec (in the initial direction of line backer)

Part B

1/2 = 0.5

KEi = 0.5*95*3.750^2 + 0.5*113*(-5.380)^2 = 2303.327 J

KEf = 0.5*(95 + 113)*(-1.210)^2 = 152.266 J

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