Question

A 77.0 kg ice hockey goalie, originally at rest, catches a 0.150 kg hockey puck slapped at him at a velocity of 22.0 m/s. Suppose the goalie and the ice puck have an elastic collision and the puck is reflected back in the direction from which it came. What would their final velocities (in m/s) be in this case? (Assume the original direction of the ice puck toward the goalie is in the positive direction. Indicate the direction with the sign of your answer.)

Answer 1

Answer:

Explanation:

For elestic collision

v₁ = $\frac{(m_1-m_2)u_1}{m_1+m_2} +\frac{2m_2u_2}{m_1+m_2}$

$v_2 = [tex]\frac{(m_2-m_1)u_2}{m_1+m_2} +\frac{2m_1u_1}{m_1+m_2}$[/tex]

Here u₁ = 0 , u₂ = 22 m/s , m₁ = 77 kg , m₂ = .15 kg ,  v₁ and v₂ are velocity of goalie and puck after the collision.

v₁ = 0 + ( 2 x .15 x22 )/ 77.15  

= .085 m / s

Velocity of goalie will be .085 m/s in the direction of original velocity of ball before collision.

v₂ = (.15 - 77)x 22 / 77.15 +0

= - 21.91 m /s

=Velocity of puck will be - 21.91 m /s  in the direction opposite  to original velocity of ball before collision.