What Is the Speed of Particle Q After the Collision?

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The discussion focuses on a collision problem involving two particles, P and Q, with masses of 2kg and 5kg, respectively. Particle P initially moves at 6m/s and reverses direction post-collision with a speed of 1.5m/s, while the goal is to find the speed and direction of particle Q after the collision. The momentum conservation equation is applied, leading to the calculation that Q's speed after the collision is 5.8m/s in the same direction it was initially moving. Additionally, there is a discussion on the impulse given to Q by P, with the impulse formula being confirmed as I = mp(v-u). The importance of correctly assigning the signs for initial velocities due to the opposite directions of motion is emphasized for accurate momentum calculations.
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Homework Statement


Two particles P and Q of mass 2kg and 5kg respectively are moving towards each other along a straight line on a smooth horizontal surface. The particles collide. Before the collision the speeds of P and Q are 6m/s and 4m/s respectively. After the collision the direction of motion of P is reversed and its speed is 1.5m/s. Find:

i) The speed and direction of Q after the collision.

Homework Equations



m1 u1 + m2 u2 = m1 v1 + m2 v2

v2 = (m1 u1 + m2 u2 – m1 v1) / m2



The Attempt at a Solution



v2 = (2 6 + 5 4 – 2 1.5) / 5

v2 = (12 + 20 – 3) / 5

v2 = 29 / 5

v2 = 5.8m/s-1

I think this equation is right and that the directiond of q would maintain if p is reversed.

ii) The magnitude of the impulse given by P to Q in the collision

i think this is the equation:

I = mp(v-u)

can someone confirm??

please
 
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If the particles are initially moving in opposite directions, their initial velocities must have opposite signs. Do they?
 
you are right, say Q in possitive and P negative!
 
nasadall said:
you are right, say Q in possitive and P negative!
And how would that change your momentum conservation equation?
 
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