Question

A 70.0-kg ice hockey goalie, originally at rest, has a 0.280-kg hockey puck slapped at him at a velocity of 45.5 m/s. suppose the goalie and the puck have an elastic collision, and the puck is reflected back in the direction from which it came. what would the final velocities of the goalie and the puck be in this case? assume that the collision is completely elastic.

Answer 1

mₐ vᵢₐ + mᵦ vᵢᵦ = mₐ vfₐ + mᵦ vfᵦ ==> [i and f means initial and final velocities] 

0.5 mₐ vᵢₐ² + 0.5 mᵦ vᵢᵦ² = 0.5 mₐ vfₐ² + 0.5 mᵦ vfᵦ² 

The collision is completely specified given the two initial velocities and the masses of the two objects. Combining the two equations above gives us a solution to the final velocities for an elastic collision of two objects: 

vfₐ = [(mₐ - mᵦ) vᵢₐ + 2 mᵦ vᵢᵦ]/[mₐ + mᵦ] 

vfᵦ = [2 mₐ vᵢₐ − (mₐ - mᵦ) vᵢᵦ]/[mₐ + mᵦ] 

substituting the values 

mₐ = 70 kg 
mᵦ = 0.280 kg 
vᵢₐ = 0 m/s 
vᵢᵦ = 45.5 m/s 

=> vfₐ = [(70 - 0.280) 0 + 2(0.280) 45.5]/[70 + 0.28] 

= 1.35 m/s ------------------->(in the original direction of the puck) 

and 

=> vfᵦ = [0 − (70 - 0.280) 45.5]/[ 70 + 0.28] = -45.14 m/s (opposite direction) 

 

 

Final speeds are therefore:

goalie = 1.35 m/s

puck = 45.14 m/s (opposite direction)

Answer 2

The final velocities of the goalie and the puck after the elastic collision are 0.4 m/s and 45.1 m/s respectively.

The given parameters:

• Mass of the goalie, m₁ = 70 kg
• Mass of the hockey puck, m₂ = 0.28 kg
• Initial velocity of the hockey puck, u₂ = 45.5 m/s

The final velocities of the goalie and the puck after the elastic collision are calculated by applying the principle of conservation of linear momentum as follows;

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\\\\70(0) + 0.28(45.5) = 70v_1+ 0.28(-v_2)\\\\12.74 = 70v_1 - 0.28v_2 \ ---(1)$

Apply one-direction velocity equation;

$u_1 + v_1 = u_2 + v_2\\\\0 + v_1 = 45.5 - v_2\\\\v_1 = 45.5- v_2 \ \ ---(2)$

Substitute the value of v1 into equation (1);

$12.74= 70(45.5 - v_2)- 0.28v_2\\\\12.74 = 3185 - 70v_2 - 0.28v_2\\\\-3172.26= -70.28v_2\\\\v_2 = \frac{3172.26}{70.28} \\\\v_2 = 45.1 \ m/s$

$v_1 = 45.5 - v_2\\\\v_1 = 45.5 - 45.1\\\\v_1 = 0.4 \ m/s$

Thus, the final velocities of the goalie and the puck after the elastic collision are 0.4 m/s and 45.1 m/s respectively.

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