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Mechanics

Q. A child of mass 48kg, is initially at rest on a stationary skateboard. The child jumps off the skateboard and initially moves horizontally with a speed of 1.2 ms^-1. The skateboard moves with a speed of 16ms-1 in the opposite direction to the direction of motion of the child. Find the mass of the skateboard. (3)

A:
48×1.2 = m×16 (2)
m = (48×1.2)/16 = 3.6kg

My question: which topic is this under and why is the answer like this? I don't understand why they're put to be as equal when it's not in equilibrium and the child is not On the skateboard either? How can they be calculate together and put as equals when the child jumped OFF the skateboard already?
Thank you so so much if anyone would help.
Original post by ChloeYeo
Q. A child of mass 48kg, is initially at rest on a stationary skateboard. The child jumps off the skateboard and initially moves horizontally with a speed of 1.2 ms^-1. The skateboard moves with a speed of 16ms-1 in the opposite direction to the direction of motion of the child. Find the mass of the skateboard. (3)

A:
48×1.2 = m×16 (2)
m = (48×1.2)/16 = 3.6kg

My question: which topic is this under and why is the answer like this? I don't understand why they're put to be as equal when it's not in equilibrium and the child is not On the skateboard either? How can they be calculate together and put as equals when the child jumped OFF the skateboard already?
Thank you so so much if anyone would help.


Conservation of momentum.

Revisit that topic and if you still dont understand this question then ask.

The speeds are given at the instant the child jumps and leaves contact with the board.
(edited 4 years ago)
Reply 2
Original post by ChloeYeo
Q. A child of mass 48kg, is initially at rest on a stationary skateboard. The child jumps off the skateboard and initially moves horizontally with a speed of 1.2 ms^-1. The skateboard moves with a speed of 16ms-1 in the opposite direction to the direction of motion of the child. Find the mass of the skateboard. (3)

A:
48×1.2 = m×16 (2)
m = (48×1.2)/16 = 3.6kg

My question: which topic is this under and why is the answer like this? I don't understand why they're put to be as equal when it's not in equilibrium and the child is not On the skateboard either? How can they be calculate together and put as equals when the child jumped OFF the skateboard already?
Thank you so so much if anyone would help.

This is a question about the conservation of momentum. When the child is on the skateboard and everything is at rest, the total momentum of the system is zero. Conservation of momentum tells us that after the child has jumped off, the total momentum will still be zero. It might have been more helpful if the first line of the answer was 48×1.2 - m×16 (2) = 0 to make this really clear. The fact that momentum is a vector, pointing in the same direction as the velocity of the object, explains why one of the final momentums is positive and the other negative.

This isn't a strictly realistic question, but the word "horizontally" is intended to suggest that all of the motion happens along a straight line.
Reply 3
Original post by Pangol
This is a question about the conservation of momentum. When the child is on the skateboard and everything is at rest, the total momentum of the system is zero. Conservation of momentum tells us that after the child has jumped off, the total momentum will still be zero. It might have been more helpful if the first line of the answer was 48×1.2 - m×16 (2) = 0 to make this really clear. The fact that momentum is a vector, pointing in the same direction as the velocity of the object, explains why one of the final momentums is positive and the other negative.

This isn't a strictly realistic question, but the word "horizontally" is intended to suggest that all of the motion happens along a straight line.

Thank you so much!! I searched the topic in my textbook but it wasn't in the index page so I searched and knew it's no longer in the new aqa spec haha. But still thank u so much for the quick and detailed reply!!
Reply 4
Original post by ChloeYeo
Thank you so much!! I searched the topic in my textbook but it wasn't in the index page so I searched and knew it's no longer in the new aqa spec haha. But still thank u so much for the quick and detailed reply!!

This makes sense - if you are doing any of the new A levels, momentum is not part of the standard A level content any more, whereas it was in all of the M1 units under the old spec. It makes it very hard to use old papers for revision!

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