Daniel B. answered 06/27/23
A retired computer professional to teach math, physics
The statement of the problem does not say anything about the angle at which the first object approaches.
I am going to assume that it approaches along the horizontal.
Without this assumption we could not solve this problem, unless we assumed conservation of energy;
that is something we cannot assume, because otherwise question b) would not be asked.
If my assumption is incorrect then please ignore this solution.
Let
m = 100g be the mass of both objects,
v = 10 cm/s be the initial velocity of the first object.
The only thing we can use is conservation of momentum.
Momentum is a vector, and both its horizontal and vertical components are conserved.
Before the collision the vertical component of the momentum is 0.
After the collision the vertical component of the momentum is
mu1sin(-30°) + mu2sin(30°).
By conservation of momentum this sum must be also 0, which implies
u1 = u2.
Let's call this common value u.
Before the collision all the momentum was in the horizontal direction, and had the magnitude
mv
After the collision the horizontal component of the total momentum is
mucos(-30°) + mucos(30°).
By conservation of momentum
mv = mucos(-30°) + mucos(30°)
Solving gives
u = v/√3 = 10/√3 = 5.77 cm/s
b)
This question calls for calculating how much kinetic energy got lost.
Before the collision the kinetic energy was
mv²/2
After the collision the total kinetic energy is
mu²/2 + mu²/2 = mu² = mv²/3
As kinetic energy is lost, this is an inelastic collision.