Chapter 5, Problem 16TE

Two football players have a head-on collision and both stop short in their paths. If one player is twice as heavy as the other, how does his speed compare to the smaller player?

To determine

The speed of the smaller player as compared heavier player.

Answer to Problem 16TE

The velocity of smaller player is twice the speed of other player.

Explanation of Solution

When two football players collide, a change in momentum occurs and there is no external force acting on them.

Write the expression for conservation of momentum.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Here,

m₁ is the mass of the first player

m₂ is the mass of the second player

u₁ is the velocity before collision of first player

u₂ is the velocity before collision of first player

v₁ is the velocity after collision of second player

v₂ is the velocity after collision of second player

Substitute 0 m/s for v₁ and v₂ in the above equation.

$$\begin{array}{c}{m}_1{u}_1+m_2{u}_2=m_1\left(0\text{m}/\text{s}\right)+m_2\left(0\text{m}/\text{s}\right)\\ \frac{u_1}{u_2}=\frac{m_2}{m_1}\end{array}$$

The mass of second player is twice heavy as first player.

Substitute 2m₁ for m₂ in the above equation.

$$\begin{array}{c}\frac{u_1}{u_2}=-\frac{2m_1}{m_1}\\ u_1=-2u_2\end{array}$$

From the above relation, it is clear that the velocity of first player having half the mass would be twice the speed of other player.

Conclusion:

Therefore, the velocity of smaller player is twice the speed of other player.