Collision of Two Particles: Conservation of Momentum and Final Velocities

AI Thread Summary
In a particle collision problem, a 5.0 kg mass moving at 8.0 m/s interacts with a stationary 6.0 kg mass. After the collision, the 5.0 kg mass moves at 4.0 m/s at an angle of 53° to the x-axis, resulting in a change in momentum of -28i -16j. The conservation of momentum principle is applied to determine the final velocity of the 6.0 kg mass, which is calculated to be 5.4 m/s. The initial momentum of the system is 40i, and the final momentum of the 5.0 kg mass is 12i - 16j, leading to the conclusion that the 6.0 kg mass has a final momentum of 28i + 16j. The discussion highlights the importance of correctly applying momentum conservation to solve collision problems.
heartshapedbox
Messages
30
Reaction score
0

Homework Statement


A particle of mass 5.0kg travels initially with a velocity of 8.0mˆı and then interacts with a particle of mass 6.0kg which was initially at rest. After the interaction the 5.0kg mass travels at a speed of 4.0m/s along a direction which makes an angle of 53◦ with the x-axis.

What is the change in momentum of the 5kg mass? -28i -16j
What is the speed of the 6kg mass after the interaction?

Homework Equations


m1v1=(m1v2)f+m2v2

The Attempt at a Solution


So I am thinking the momentum must be conserved. If the initial velocity was 8m/s in the i direction, the final velocity must be as well. The final velocity of the 5kg mass in component form is 4cos(307)=2.4i and 4sin(307)=-3.2j. This means the final velocity of the 6kg mass is 5.6i and 3.2j, giving 6.5 as the final velocity.

Apparently the answer is 5.4, can you show me what i should be doing instead. Thanks :)
 
Physics news on Phys.org
heartshapedbox said:
So I am thinking the momentum must be conserved. If the initial velocity was 8m/s in the i direction, the final velocity must be as well.
These two statements, don't mean the same thing. How did you conclude that the final velocity must equal the initial?
 
Ok i see what i did wrong, lol, long day.
Total initial momentum: 40i
5kg mass final momentum: 12i, -16j
Due to conservation of momentum: 6kg final momentum: 28i, 16j
sqrt((28/6)^2+(-16/6)^2)
final speed: 5.4m/s
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Back
Top