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Consider a problem as follows.

A dog with a mass of 5 kg is at rest on a moving boat with mass of 20 kg. The speed of the boat is 1 m/s (relative to the ground). If the dog leaves the boat by jumping at the speed of 2 m/s (relative to the boat) in the direction of the boat moves, find the speed of the boat (relative to the ground) after this event? Ignore any possible frictional forces.

I am not sure that my calculation below is correct. I take the ground as the frame of reference.

Before Jumping

Before jumping both dog and boat have the same speed relative to the ground, i.e., 1 m/s. So momentum of the system before jumping is

\begin{align} p &=m_dv_d+m_bv_b \\ &=5\times1+20\times1=25\\ \end{align}

After Jumping

When the dog jumps at speed 2 m/s relative to the boat, the boat moves at speed $v_b'$ relative to the ground. It means the dog jump at speed $v_d'=2+v_b'$ relative to the ground.

Momentum of the system after jumping is \begin{align} p' &=m_dv_d'+m_bv_b'\\ &=5\times(2+v_b')+20\times v_b'\\ &=10+25v_b' \end{align}

Applying the law of conservation, I have

\begin{align} p&=p'\\ 25&=10+25v_b'\\ v_b'&=0.6 \end{align}

But according to the answer key, $v_b'=1/2$ m/s.

Questions

Which is the correct answer? mine or the answer key?

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closed as off-topic by John Rennie, Wolpertinger, user36790, ACuriousMind, Jon Custer Oct 2 '16 at 16:18

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velocity of dog after jumping from the boat is 3m/s, since it already had a velocity of 1m/s and another 2m/s relative to the boat. Change the expression of velocity of dog wrt ground as $v_{dg}=v_{db}+v_{bg}$ (velocity of dog wrt ground= velocity of dog wrt boat+velocity of boat wrt ground)

Initial momentum = final momentum in the direction in which there is no force....

$(5+20)$($1m/s$) = $5$($3m/s$) + $20$*($x$ $m/s$)

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  • $\begingroup$ Please write the complete equation. $\endgroup$ – Artificial Stupidity Oct 2 '16 at 13:38
  • $\begingroup$ After jumping, the boat speed changes to $x$ so the dog speed relative to the ground become $2+x$ rather than 3, right? $\endgroup$ – Artificial Stupidity Oct 2 '16 at 14:13
  • $\begingroup$ the dog did not jump with $2m/s$ relative to the boat when the boat's speed was $x$, it jumped when the boat's speed was $1$. so it is $2 + 1$ $\endgroup$ – Prasad Mani Oct 2 '16 at 14:26
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Your answer is wrong as ,if you are performing all calculations w.r.t. ground then you also need to take the velocity of dog w.r.t. ground which would be 3m/s as the boat is already moving with a velocity of 1m/s w.r.t. ground and dog has jumped from the boat with velocity 2m/s w.r.t. the boat therefore the velocity of dog w.r.t. ground will be 3m/s , if you take this value in your calculations then you will get the velocity of boat as 1/2 m/s .

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