Question

Two boys are sliding toward each other on a frictionless, ice-covered parking lot. Jacob, mass 45 kg, is gliding to the right at 8.08 m/s, and Ethan, mass 31.0 kg, is gliding to the left at 10.9 m/s along the same line. When they meet, they grab each other and hang on.(a) What is their velocity immediately thereafter?

Answer 1

Answer:

$v=0.3381\ m.s^{-1}$ in the direction of motion of Jacob

Explanation:

Given:

mass  of Jacob, $m_j=45\ kg$

velocity of Jacob, $v_j=8.08\ m.s^{-1}$

mass of Ethan, $m_e=31\ kg$

velocity of Ethan, $v_e=10.9\ m.s^{-1}$

Now using the conservation of linear momentum for the case:

(When the two masses in motion combine to form one after the collision then they will move together in the direction of the greater momentum.)

$(m_j+m_e).v=m_j.v_j-m_e.v_e$

$(45+31)\times v=45\times 8.08-31\times 10.9$

$v=0.3381\ m.s^{-1}$ in the direction of motion of Jacob as it was assumed to be positive.