Question

You and your friends are doing physics experiments on a frozen pond that serves as a frictionless, horizontal surface. Sam, with mass 73.0 kg, is given a push and slides eastward. Abigail, with mass 57.0 kg, is sent sliding northward. They collide, and after the collision Sam is moving at 40.0^\circ north of east with a speed of 7.00 m/s and Abigail is moving at 18.0^\circ south of east with a speed of 9.40 m/s. What was the speed of each person before the collision?

a. Sam's speed:
b. Abigail's speed:
c. By how much did the total kinetic energy of the two people decrease during the collision?

Answer 1

Sam's and Abigail speeds before colliding were a. 12.34 m/s and b. 2.86 m/s, respectively. Their total kinetic energy was diminished by c. 1484.42 J, approximately

Step-by-step explanation:

By conservation of momentum, we have

`\Delta \vec{p} = 0\\\\m\vec{v}_i = m\vec{v}_f\\m_S v_S\hat{i} + m_A v_A\hat{j} = m_S * 7 cos((40)) \hat{i} + m_S * 7 sin((40)) \hat{j} + m_A * 9.4 cos((18)) \hat{i} - m_A * 9.4 sin((18)) \hat{j}.`

Writing for each direction at a time,

`m_S v_S = 7m_S cos((40)) +9.4 m_A cos((18))\\v_S = 7 cos((40)) + 9.4 (m_A)/(m_S) cos((18)) = 7 cos((40)) + 9.4*(57)/(73) cos((18)) \approx \mathbf{12.34}\\m_A v_A = 7m_Ssin((40)) - 9.4m_Asin((18))\\v_A = 7(m_S)/(m_A)sin((40)) - 9.4sin((18)) = 7*(73)/(57)sin((40)) - 9.4sin((18)) \approx \mathbf{2.86}.`

Their kinetic energy changed by

`K_f - K_i = \left( (1)/(2)m_s v_(fs)^2 + (1)/(2)m_a v_(fa)^2 \right) - \left( (1)/(2)m_s v_(is)^2 + (1)/(2)m_a v_(ia)^2 \right) =  \left( (1)/(2)* 73 * 7^2 + (1)/(2)* 57 * 9.4^2 \right) - \left( (1)/(2)* 73 * 12.34^2  + (1)/(2)* 57 * 2.86^2 \right) \approx \mathbf{-1484.418 J}.`