Question

Running at 2 m/s, Bruce the 45 kg quarterback, collides with Biff, the 90 kg tackle, who is traveling at 7 m/s in the other direction. Upon collision, Biff continues to travel forward at 1 m/s. How fast is Bruce knocked backwards? What is Bruce’s impulse?

Answer 1

Given data

*The mass of Bruce is m_1 = 45 kg

*The initial velocity of the Bruce is u_1 = 2 m/s

*The mass of the biff is m_2 = 90 kg

*The initial velocity of the Biff is u_2 = -7 m/s

*The final velocity of the first glider is v_2 = -1 m/s

According to the law of conservation of linear momentum, the total linear momentum of a system remains constant

Applying the law of conservation of momentum as

$\begin{gathered} p_i=p_f \\ m_1u_1+m_2u_2=m_1v_1+m_2v_2 \\ v_1=\frac{m_1u_1+m_2u_2-m_2v_2_{}_{}_{}_{}}{m_1} \end{gathered}$

Substitute the known values in the above expression as

$\begin{gathered} v_1=\frac{(45)(2)+(90)(-7)-(90)(-1)}{45} \\ =-10\text{ m/s} \end{gathered}$

Hence, the speed of the bruce knock backwards is v_1 = -10 m/s