How Is the Change in Linear Momentum Calculated in a Collision?

[ViewtifulBeau]

A 0.118 kg mass is moving horizontally with a speed of 8.49 m/s when it strikes a vertical wall. The mass rebounds with a speed of 1.07 m/s. What is the magnitude of the change in linear momentum of the mass?

isn't the change just the difference in momentum? I did .118(8.49 - 1.07) to get .87556 kg m/s. this is wrong so i think the answer is .118(8.49+1.07) because it would be minus a negative velocity. so i got 1.128 kg m/s. but i don't know if this number should be negative or not.

 

[Fermat]

The change is, as you say, just the differnce in momentum.

But momentum is mv. And v (the velocity) is a vector and has a direction.
So deltaP = mv1 - mv2, but v1 and v2 are in opposite directions, so you have to (numerically) add them.

Since you are asked for a differnce, then take the absolute value, i.e. positive.

 

[quicknote]

I can confirm that the magnitude of the change in linear momentum of the mass is 1.128 kg m/s. The change in momentum is calculated by taking the difference between the initial momentum (m x v) and the final momentum (m x v'). In this case, the initial momentum is 0.118 kg x 8.49 m/s = 1.00042 kg m/s and the final momentum is 0.118 kg x 1.07 m/s = 0.12606 kg m/s. Taking the difference between the two gives us 1.00042 kg m/s - 0.12606 kg m/s = 0.87436 kg m/s. This is the negative change in momentum, indicating that the mass lost momentum in the collision with the wall. To get the magnitude, we take the absolute value of this number, which is 0.87436 kg m/s. However, since momentum is a vector quantity, the magnitude should also have a direction. In this case, the direction of the change in momentum is opposite to the initial momentum, indicating that the mass rebounded in the opposite direction. Therefore, the complete answer is that the magnitude of the change in linear momentum is 0.87436 kg m/s in the opposite direction to the initial momentum.