Answer:
s₁ = 0.022 m
Explanation:
From the law of conservation of momentum:
[tex]m_1u_1 + m_2u_2 = m_1v_1+m_2v_2[/tex]
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,
[tex](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[/tex]
negative sign here shows the opposite direction.
Now, we calculate the time taken by puck to move 14.5 m:
[tex]s_2 =v_2t\\\\t = \frac{s_2}{v_2} = \frac{14.5\ m}{48\ m/s} \\\\t = 0.3\ s[/tex]
Now, the distance covered by the player in this time will be:
[tex]s_1 = v_1t\\s_1 = (0.074\ m/s)(0.3\ s)[/tex]
s₁ = 0.022 m
