a body of mass m collides with a wall (angle between ball and wall is 60 degrees) if the speed (v)doesnt change when rebounding...............find its change in momentum......HELP PLS!!!!!!!!

No change in speed.

Only the direction of velocity changes.

So collides at angle 60 deg to wall, and also rebounds at angle 60 deg to wall, like a "V".

So change in momentum = 2vsin60*m = √3mv. (only along the vertical direction - momentum doesn't change along the horizontal direction).

Hope that helps.

Best Regards,

Ashwin (IIT Madras).

break d momentum in components....at horizontal mv cos 60 at vertical mv sin 60....the chnge in momnetum will be 0 in vertical direction but it will chnge its path after collision in horizonntal direction....so change is mv/2 -(-mv/2) = mv

When the ball of mass m collides with the ball with velocity v₁ making angle 30° with the normal to the wall,it rebounds with velocity v₂ at an angle 30° to the normal on the wall (angle between the ball and the normal=90°-angle between the wall and the ball).Here,the speedc v₁ and v₂ are same but the directions are different,so the velocities are different.

now,change in momentum would be

Δp=mv₁-mv₂

Δp=m(v₁-v₂)

Subtracting the two velocities vectorially,we get,

Δp=m[√(v₁²+v₂²-2v₁v₂cosθ)],here θ is the angle between the two velocities.

Δp=m[√(v²+v²-2v²cos120°)]

Δp=m[√(2v²-2v²(-1/2))]

Δp=m[v√3]

so,the change in momentum is (mv√3).