Question

An 70 kg running back has a football and is making a break for a touchdown at 9 m/s. On his way, a 126 kg linebacker running in the opposite direction at 4 m/s collides with him for the tackle. How much kinetic energy is lost in this collision?

Round intermediate answers to four significant figures and final answers to three significant figures for best accuracy.

Answer 1

This is the case of one dimensional inelastic collision. So after the collision, both the players will be moving with the same velocity.

Let M₁ be the mass of the first player and V_1,i be the initial velocity of the first player. Similarly M₂ be the mass of the second player and V_2,i be the initial velocity of the second player. We fill take the direction in which player 1 is moving as the positive direction of velocity.

M₁ = 70 kg, V_1,i = 9 m/s

M₂ = 126 kg, V_2,i = -4 m/s ( because it is moving in the opposite direction to player 1)

The total kinetic energy before the collision is:

K.E._i = (1/2)M₁V_1,i² + (1/2)M₂V_2,i²

Putting all the values we get

K.E._i = (1/2)*70*9*9 + (1/2)*126*(-4)*(-4)

        = 2835 + 1008

K.E._i = 3843 J

Now we have to find the final velocity with which both the players are moving.

Applying conservation of Linear momentum (Because there is no external force on the system)

     M₁V_1,i + M₂V_2,i = (M₁ + M₂)V

where V is the final velocity of both the players.

Putting all the values we get

    70*9 + 126*(-4) = (70 + 126)V

    630 - 504 = 196*V

    126 = 196*V

Dividing both sides by 196 we get

   126/196 = V

   V = 0.6428571 m/s

After rounding off to four significant figures, it becomes

V = 0.6429 m/s

The final kinetic energy of the system is:

K.E._f = (1/2)*(M₁ + M₂)*V²

Putting all the values we get

K.E._f = (1/2)*(70 + 126)*(0.6429)*(0.6429)

K.E._f = 40.5054 J

After rounding off to four significant figures, it becomes

K.E._f = 40.51 J

The amount of kinetic energy lost is equal to:

K.E._lost = K.E._i - K.E._f

Putting all the values we get

K.E._lost = 3483 - 40.51

K.E._lost = 3442.49 J

After rounding off to three significant figures, it becomes

K.E._lost = 344*10¹ J

So a total of 344*10¹ J of energy is lost in the collision.