elastic collisions.....?

Question

Two objects of masses m and 4m are moving toward each other along the x-axis with the same initial speeds v0 = 45.0 m/s. The object with mass m is traveling to the left, and the object with mass 4m is traveling to the right. They undergo an elastic glancing collision such that mass m is moving downward after the collision at right angles from its initial direction.
(a) Find the final speeds of the two masses.
(b) What is the angle from its initial direction of motion at which the mass 4m is scattered?

Answer 1

Elastic collision means that both momentum AND energy are conserved.

take right motion and upward as positive.

4*m*45-m*45=4m*vx

solve for vx
3*45/4=vx
vx=33.75 m/s

.5*4*m*45^2+.5*m*45^2=.5*4*m*v^2+.5*m*vy^2

4*45^2+45^2=4*v^2+vy^2

we know that v^2=vy^2+vx^2
or
v^2=vy^2+33.75^2
so
4*45^2+45^2=4*(vy^2+33.75^2)+vy^2

solve for vy
vy=33.37 m/s

so the mass m is moving downward with a speed of 33.37 m/s and the mass 4m is moving at a speed of
sqrt(33.37^2+33.75^2)
or
47.46 m/s

the angle is
atan(33.37/33.75)
or 44.7 degrees above horizontal.

j