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In a typical Indian Bugghi (a luxury cart drawn by horses), a wooden plate is fixed on the rear on which one person can sit. A Bugghi of mass $200 kg$ is moving at a speed of $10 km/h$. As it overtakes a schoolboy walking at a speed of $4 km/h$, the boy sits on the wooden plate. If the mass of the boy is $25 kg$, what will be the new velocity of the Bugghi?

Well intuitively by using basic momentum concept, I solved it like this, $$(200 \text{ kg} )(\frac{10000}{3600}\text{ m/s})=(250 \text{ kg})(v^{'})$$ and found the value of $v^{'}$. This means that the momentum of the boy is lost when he is in the process of sitting. This I thought was logical because when the boy boards the Bugghi, his velocity will become zero anyways, when he seats on it.

However, according to the answer sheet of this problem, the total momentum before the process of sitting remains the same after the process of sitting.

How is this possible as the boy sits on the Bugghi his velocity is zero and he should have no momentum by himself when on the Bugghi?

Please explain this small understanding error I might have had here. Thank You!

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Firstly the boy doesn't come to rest when he boards the bugghi; he and the bugghi start moving with the same speed.

Wrt the bugghi the boy is at rest, but in the ground frame both have the same speed.

Secondly, momentum of the boy is not conserved, as there is external force on him. When he sits on the bugghi frictional force helps him not slip. In the same way the momentum of the bugghi is also not conserved as it experiences friction due to the boy.

But if we take both of them together as our system, we observe that these frictional forces are all inside our system, which should not affect the collective momentum of the system.

So the momentum of the SYSTEM is what you need to conserve, not of the man or the bugghi.

Hope it helped.

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The answer sheet is stating that momentum is conserved ($\Delta p_{system} = 0$). This means $p_{\small initial} = p_{\small final}$. Momentum is conserved when no external forces are acting on the "system" comprised of the boy and the cart (and the horses!).

Yes, in reality there could be wind resistance or friction or any other innumerable forces acting but we are ignoring those forces and assuming that none of them are acting on the boy, cart or horses (with the technicality that static friction is needed for the boy to run and the wheels to turn to move the cart and the horses to jog).

The velocity of the boy doesn't become zero when he sits on the cart. His velocity becomes the same as the cart. The $v'$ you are trying to solve for is the velocity the cart changes to once the boy sits on it. Obviously if the cart is moving at $v'$ and the boy is on the cart then the boy must be moving at $v'$ too.

While you rightly combined the boy with the cart as a single moving object after the boy sat on the cart, you didn't include the momentum of the boy prior to sitting on the cart. The boy has his own momentum while walking/running. $$p_{\small initial} = p_{\small boy} + p_{\small cart}$$

With this included you can calculate :

$$p_{\small final} = p_{\small \,boy \,+ \,cart}$$

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In this question, take the boy, the Bugghi (along with the person in it), the horses drawing it, and the ground as the system. There are no external forces, so momentum must be conserved.

Your error lies in assuming that the boy will move with zero velocity after sitting on the Bugghi. His velocity relative to the Bugghi is zero, but it is non-zero relative to the ground. Hence, both the boy and the Bugghi will move at the same velocity after he sits on it.

You must be careful about which reference frames you're applying conservation of momentum in, and not switching them in two sides of the same equation.

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    $\begingroup$ Are not the horses exerting an external force on the system as defined? $\endgroup$ – DJohnM Apr 27 at 6:06
  • $\begingroup$ Good catch, thank you. I'll add that the horses are a part of the system. $\endgroup$ – wavion Apr 27 at 6:07
  • $\begingroup$ But then there is an external horizontal force from the ground on the horses' hooves, so no conservation of linear momentum for the new system... $\endgroup$ – DJohnM Apr 27 at 17:11
  • $\begingroup$ I included the ground in the system too, thanks a lot for the careful critique :) $\endgroup$ – wavion Apr 28 at 16:34

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