You write that:
"(A)fter the collision, both ball(s) move at 30 degree and 15 degree from horizontal. "
Which is it? Why use the word "both" if they leave at different angles?
This is an exercise in momentum and kinetic energy conservation.
Since there are only two unknowns, and there are two momentum equations, you should not have to assume kinetic energy conservation to solve the problem.
A 2kg ball,A moves to the right with velocity 4 m/s collides with another 1kg ball,B moving in the opposite way with velocity 0.5m/s. after the collision, both ball move at 30 degree and 15 degree from horizontal. calculate the final velocity of each ball after the collision. assume the collision is elastic.
3 answers
ball A move 30 degree from horizontal while ball B move 15 degree from horizontal.
x-momentum conservation:
2*4 - 1*(1/2) = Va*cos30 + Vb*cos15
y-momentum conservation:
0 = Va*sin30 - V2*sin15
7.5 = 0.866*Va + 0.966*Vb
0 = 0.500*Va - 0.259*Vb
Vb = 1.93 Va
Substitution will let you solve for Va.
7.5 = 0.866Va + 1.865Va = 2.731 Va
Va = 2.75 m/s
Vb = 5.30 m/s
Initial KE = (1)*4^2 + (1/2)(0.5)^2 = 16.25 J
Final KE = (1)*2.75^2 + (1/2)*(5.30)^2
21.61 J
The situation described cannot happen. In order to conserve momentum and have those directions, the collision must be superelastic
2*4 - 1*(1/2) = Va*cos30 + Vb*cos15
y-momentum conservation:
0 = Va*sin30 - V2*sin15
7.5 = 0.866*Va + 0.966*Vb
0 = 0.500*Va - 0.259*Vb
Vb = 1.93 Va
Substitution will let you solve for Va.
7.5 = 0.866Va + 1.865Va = 2.731 Va
Va = 2.75 m/s
Vb = 5.30 m/s
Initial KE = (1)*4^2 + (1/2)(0.5)^2 = 16.25 J
Final KE = (1)*2.75^2 + (1/2)*(5.30)^2
21.61 J
The situation described cannot happen. In order to conserve momentum and have those directions, the collision must be superelastic
2 answers