Question

Two 22.7 kg ice sleds A and B are placed a short distance apart, one directly behind the other, as shown in fig. A 3.63 kg dog, standing on one sled, jumps across to the other and immediately back to the first. Both jumps are made at a speed of 3.05 ms! relative to the ice. Find the final speeds of the two sleds.​

Answer 1

Given that :-

The masses of the two ice sleds, $m_{1} = m_{2} = 22.7 kg$

The initial speed of the ice, $v_{1} =v_{2}=0$

The mass of the cat, $m_{3} = 3.63 kg$

The initial speed of the cat, $v_{3} =0$

The horizontal speed of the cat, $v_{3}=3.05m/v$

As we have to find the two sleds final speed,

 For first sleds,

By using the law of conservation of momentum,  

Initial momentum = Final momentum

∴ $m_{1}$ × $v_{1} + m_{3}$ × $v_{3}=m_{1}$ × $v^{'} _{1} + m_{3}$ × $v^{'} _{2}$

Now, by plugging in the values gives;

∴  $22.7 kg$× $0 + 3.63$ × $0 = 22.7$ × $v^{'} _{1} +3.63$ × $3.05$

∴ $22.7$× v₁'  $= -3.63$×$3.05$

v₁' = $-3.63$ × $\frac{3.05}{22.7}$ ≈ $-0.49$

 As the final speed ≈ $0.49 m/s$

Similarly, for second sleds,

We use conservation of momentum law,

 Again substituting the values,

∴ $22.7 kg *$$0 + 3.63 * 0 = 22.7 * v^{'}_{2}$ $+ 3.63 * 3.05$

  $v^{'}_{2}=-3.63 * \frac{3.05}{22.7}$

$v^{'}_{2}=-0.49$

 Thus, the final speed ≈ 0.49 m/s