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3 years ago

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A 97.0 kg ice hockey player hits a 0.150 kg puck, giving the puck a velocity of 48.0 m/s. If both are initially at rest and if t

he ice is frictionless, how far does the player recoil in the time it takes the puck to reach the goal 14.5 m away

1 answer:

OverLord2011 [107]3 years ago

4 0

Answer:

s₁ = 0.022 m

Explanation:

From the law of conservation of momentum:

$m_1u_1 + m_2u_2 = m_1v_1+m_2v_2$

where,

m₁ = mass of hockey player = 97 kg

m₂ = mass of puck = 0.15 kg

u₁ = u₂ = initial velocities of puck and player = 0 m/s

v₁ = velocity of player after collision = ?

v₂ = velocity of puck after hitting = 48 m/s

Therefore,

$(97\ kg)(0\ m/s)+(0.15\ kg)(0\ m/s)=(97\ kg)(v_1)+(0.15\ kg)(48\ m/s)\\\\v_1 = -\frac{(0.15\ kg)(48\ m/s)}{97\ kg} \\v_1 = - 0.074 m/s$

negative sign here shows the opposite direction.

Now, we calculate the time taken by puck to move 14.5 m:

$s_2 =v_2t\\\\t = \frac{s_2}{v_2} = \frac{14.5\ m}{48\ m/s} \\\\t = 0.3\ s$

Now, the distance covered by the player in this time will be:

$s_1 = v_1t\\s_1 = (0.074\ m/s)(0.3\ s)$

<u>s₁ = 0.022 m</u>

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