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irina [24]
2 years ago
10

A 0.105- kg hockey puck moving at 24 m/s is caught and held by a 75-kg goalie at rest. With what speed does the goalie slide on

the ice after catching the puck? Define the puck and the goalie as a system.
please show work, help
Physics
1 answer:
Veseljchak [2.6K]2 years ago
3 0

Answer:

<u><em>0.03 m/s</em></u>

Explanation:

<em>Applying law of conservation of momentum, </em>

  • <em>m₁v₁ + m₂v₂ = (m₁ + m₂)v</em>
  • <em>0.105(24) + 75(0) = (0.105 + 75)v</em>
  • <em>75.105v = 2.52</em>
  • <em>v = 2.52/75.105</em>
  • <em>v = </em><u><em>0.03 m/s</em></u>
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Now, the length is shortened by 1.0 m. So, the new length is 1 m less than the original length.

New length of the pendulum is, L_1=L-1

New time period of the pendulum is, T_1=2.81\ s

We know that, the time period of a simple pendulum of length 'L' is given as:

T=2\pi\sqrt{\frac{L}{g}}-------------- (1)

So, for the new length, the time period is given as:

T_1=2\pi\sqrt{\frac{L_1}{g}}------------ (2)

Squaring both the equations and then dividing them, we get:

\dfrac{T^2}{T_1^2}=\dfrac{(2\pi)^2\frac{L}{g}}{(2\pi)^2\frac{L_1}{g}}\\\\\\\dfrac{T^2}{T_1^2}=\dfrac{L}{L_1}\\\\\\L=\dfrac{T^2}{T_1^2}\times L_1

Now, plug in the given values and calculate 'L'. This gives,

L=\frac{3.45^2}{2.81^2}\times (L-1)\\\\L=1.507L-1.507\\\\L-1.507L=-1.507\\\\-0.507L=-1.507\\\\L=\frac{-1.507}{-0.507}=2.97\ m

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