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 The variation of the object's distance d with time t is shown on the graph below.

In the case of an object's movement resembled in this graph, as the gradient is decreasing, a decrease in velocity occurs.

According to the formula

$$p = mv$$,

The momentum is directly proportional to the velocity of the object, leading it to decrease in the case shown above in the graph, as the velocity decreases.

The question is: How does the direction of the change in momentum of the object change during the motion. (The model answer says that since the final momentum is less than the initial momentum - which is right -, the result of the difference in momentum (∆p) should be negative - which is also right -, indicating that the direction of the change in momentum is opposite to the motion of the object. Leading the direction of the momentum to be opposite to the motion/backwards/negative.)

How come this and the direction of the velocity was never opposite the reference point, i.e., the velocity's direction in the period shown in the graph is never downwards for example.

Isn't all what happened for the velocity just a change in magnitude, how come this anyway influence the direction of momentum.

If there's a point I'm not understanding, please clarify, or in case my talk is logical, tell me wether the mark scheme is inaccurate.

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  • $\begingroup$ Assuming that this is about one dimensional motion and $d$ is the displacement then the gradient of the graph which is always positive gives the velocity. Since the gradient decreases with time then so does the velocity and the momentum and thus the change in momentum is negative. $\endgroup$
    – Farcher
    Commented Jul 1 at 9:35
  • $\begingroup$ @Farcher, I got your point, but does a negative change in momentum indicates a change in the direction of momentum. The simple answer is no. So how such a negative change in magnitude relate to a change in the direction? $\endgroup$ Commented Jul 1 at 10:06
  • $\begingroup$ There is no change in direction. The "change in momentum", in this graph, is always opposite of that of its velocity, every "where" and every "when" of this graph (as drawn). $\endgroup$ Commented Jul 1 at 10:11
  • $\begingroup$ I stated that I assumed it was one dimensional motion and hence there are only two directions possible, in a direction designated positive, and in an opposite direction which would be designated negative. $\endgroup$
    – Farcher
    Commented Jul 1 at 14:16

1 Answer 1

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  1. You're correct that the velocity's direction doesn't change in this graph. The object is always moving in the positive direction (upwards on the distance-time graph), just slowing down.

  2. The change in momentum (Δp) being negative doesn't mean the momentum itself is negative. It just means the final momentum is less than the initial momentum.

  3. The direction of the change in momentum is indeed opposite to the motion, but this doesn't mean the object is moving backwards. It means the force causing the change is acting in the opposite direction of the motion.

To illustrate:

  • The object starts with a high velocity and momentum.
  • A force acts on it in the opposite direction of its motion, causing it to slow down.
  • This force results in a change in momentum that's opposite to the motion.
  • The object's velocity and momentum decrease but remain positive.

The mark scheme is correct, but it's easy to misinterpret. The key is distinguishing between: a) The direction of motion (always positive here) b) The direction of the change in momentum (negative, opposing the motion)

Your understanding that the velocity only changes in magnitude is correct. The direction of the change in momentum refers to the direction of the force causing the change, not the direction of the momentum itself.

All in all, your reasoning is logical, and you've identified a subtle but important distinction. The mark scheme is accurate, but it requires careful interpretation to avoid confusion between the direction of motion and the direction of the change in momentum.

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