Before After m1=1.0 kg m2=2.0 kg m1=1.0 kg m2=2.0 kg v1=2.0 m/s v2=−1 m/s v1=−2.0 m/s v2=1 m/s
Given the data, what is the kinetic energy of the system before and after in this elastic collision?
A -1J
B 3J
C 0J
D 6J
To calculate the kinetic energy before and after the collision, we can use the formula:
KE = (1/2) * mass * velocity^2.
Before the collision:
KE1 (m1) = (1/2) * 1 kg * (2.0 m/s)^2 = 2 J
KE2 (m2) = (1/2) * 2 kg * (-1 m/s)^2 = 1 J
Total KE before the collision = KE1 + KE2 = 2 J + 1 J = 3 J.
After the collision, we need to calculate the final velocities using the conservation of momentum:
m1v1 + m2v2 = m1u1 + m2u2,
where u is the final velocity.
Using the given data,
m1v1 + m2v2 = m1u1 + m2u2,
(1 kg * -2 m/s) + (2 kg * 1 m/s) = (1 kg * u1) + (2 kg * u2),
-2 + 2 = u1 + 2u2,
0 = u1 + 2u2.
We have two equations now:
1. m1v1 + m2v2 = m1u1 + m2u2,
2. 0 = u1 + 2u2.
Solving these equations simultaneously, we find:
u1 = 2 m/s
u2 = -1 m/s.
After the collision:
KE1 (m1) = (1/2) * 1 kg * (2.0 m/s)^2 = 2 J
KE2 (m2) = (1/2) * 2 kg * (-1 m/s)^2 = 1 J
Total KE after the collision = KE1 + KE2 = 2 J + 1 J = 3 J.
Therefore, the kinetic energy of the system before and after the elastic collision is 3J. Therefore, the answer is B.