Calculating change of momentum during an impactWatch

Announcements
Thread starter 1 week ago
#1
What is the formula for calculating the change of momentum.

I am given the mass, the speed, how long the impact lasted for and how much speed it caused it to increase. How do I work out the change in momentum of the impact? what is the formula
0
1 week ago
#2
Momentum is mass multiplied by velocity (speed) - so the change in momentum is (mass x speed) afterwards minus the (mass x speed) before. So if the mass is 10kg and the speed before is 10m/s, and the speed after is 12m/s then, mass x speed after = 10x12=120 Kg m/smass x speed before = 10x10=100 Kg m/sSo change in momentum is 120-100=20 Kg m/sGood luck!
1
Thread starter 1 week ago
#3
(Original post by SurfingSam)
Momentum is mass multiplied by velocity (speed) - so the change in momentum is (mass x speed) afterwards minus the (mass x speed) before. So if the mass is 10kg and the speed before is 10m/s, and the speed after is 12m/s then, mass x speed after = 10x12=120 Kg m/smass x speed before = 10x10=100 Kg m/sSo change in momentum is 120-100=20 Kg m/sGood luck!
thank you. What happens if I am given no initial velocity. For example I have this question:

A molecule of mass 5.0*10-26kg moving at a speed of 420ms-1 hits a surface at right angles to the surface and rebounds at the same speed in the opposite direction in an imoact lasting 0.22ns
I'm given the mass and speed, but not the initial speed. How do I work it out?
0
1 week ago
#4
(Original post by Simon33355)
thank you. What happens if I am given no initial velocity. For example I have this question:

A molecule of mass 5.0*10-26kg moving at a speed of 420ms-1 hits a surface at right angles to the surface and rebounds at the same speed in the opposite direction in an imoact lasting 0.22ns
I'm given the mass and speed, but not the initial speed. How do I work it out?
The size of the velocity (ie the speed) is the same before and after (420m/s) but the direction has changed. Before it's was heading towards the surface (let's say in a positive direction), after it's heading away from the surface (so in a negative direction).

The change in momentum is therefore (mass x velocity) before minus (mass x velocity after)
This is (5.0*10-26 x 420) - (5.0*10-26 x -420)
That's the same as (5.0*10-26 x 420) + (5.0*10-26 x 420) because the two minuses make a plus
Good luck again!
0
Thread starter 1 week ago
#5
(Original post by SurfingSam)
Momentum is mass multiplied by velocity (speed) - so the change in momentum is (mass x speed) afterwards minus the (mass x speed) before. So if the mass is 10kg and the speed before is 10m/s, and the speed after is 12m/s then, mass x speed after = 10x12=120 Kg m/smass x speed before = 10x10=100 Kg m/sSo change in momentum is 120-100=20 Kg m/sGood luck!
thank you. What happens if I am given no initial velocity. For example I have this question:

A molecule of mass 5.0*10-26kg moving at a speed of 420ms-1 hits a surface at right angles to the surface and rebounds at the same speed in the opposite direction in an imoact lasting 0.22ns
I'm given the mass and speed, but not the initial speed. How do I work it out?
0
X

new posts Back
to top
Latest
My Feed

Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

University open days

• Cardiff Metropolitan University
Undergraduate Open Day - Llandaff Campus Undergraduate
Sat, 19 Oct '19
• Coventry University
Sat, 19 Oct '19
• University of Birmingham
Sat, 19 Oct '19

Poll

Join the discussion

Why wouldn't you turn to teachers if you were being bullied?

They might tell my parents (20)
6.73%
They might tell the bully (30)
10.1%
I don't think they'd understand (44)
14.81%
It might lead to more bullying (111)
37.37%
There's nothing they could do (92)
30.98%