How Does Conservation of Momentum Apply in 2D Elastic Collisions?

[HelloMotto]

i seem to having trouble with questions related to elastic collisions. Please help me out if you can please.

A 0.25 kg ball is attached to a 26-cm piece of string. The ball is first raised so that the string is taut and horizontal, then the ball is released so that, at the bottom of its swing, it undergoes an elastic headon collision with a 0.21 -kg ball that is free to roll along a horizontal table.

What is the speed of the 0.21 kg ball just after the collision?

 

[olgranpappy]

i seem to having trouble with questions related to elastic collisions. Please help me out if you can please.

A 0.25 kg ball is attached to a 26-cm piece of string. The ball is first raised so that the string is taut and horizontal, then the ball is released so that, at the bottom of its swing, it undergoes an elastic headon collision with a 0.21 -kg ball that is free to roll along a horizontal table.

What is the speed of the 0.21 kg ball just after the collision?

first figure out the speed of the 0.25kg ball right before the collision.

 

[HelloMotto]

first figure out the speed of the 0.25kg ball right before the collision.

ok i got like 2.26 m/s just before the collision.
i know that in perfectly elastic collision, the kinetic energy is conserved.
so my energy statement would be something like this
Ek initial = Ek ball1 + Ek ball2. right?

but when i expand and plug in the givens, I am left with 2 unknowns, the final velocity of each ball. Thats where I am stuck at.

 

[olgranpappy]

ok i got like 2.26 m/s just before the collision.
i know that in perfectly elastic collision, the kinetic energy is conserved.
so my energy statement would be something like this
Ek initial = Ek ball1 + Ek ball2. right?

but when i expand and plug in the givens, I am left with 2 unknowns, the final velocity of each ball. Thats where I am stuck at.

the solution to your problem is given as the title to this thread... use conservation of momentum.