3. A 10.0-g marble slides to the left with a velocity of magnitude 0.4m/s on a frictionless, horizontal surface and has a head-on, elastic collision with a larger 30.0-g marble sliding to the right with a velocity of 0.2m/s magnitude.
(a) Find the velocity of each marble (magnitude and direction) after the collision. (b) Calculate the change in momentum.
(a) To find the velocity of each marble after the collision, we can use the conservation of momentum and conservation of kinetic energy principles.
Let v1 be the velocity of the smaller marble after the collision and v2 be the velocity of the larger marble after the collision.
Conservation of momentum:
m1u1 + m2u2 = m1v1 + m2v2
10g * 0.4m/s + 30g * (-0.2m/s) = 10g * v1 + 30g * v2
4g - 6g = 10g * v1 + 30g * v2
-2g = 10g * v1 + 30g * v2
-2 = 10v1 + 30v2
-2 = 10v1 + 30v2
Conservation of kinetic energy:
0.5m1u1^2 + 0.5m2u2^2 = 0.5m1v1^2 + 0.5m2v2^2
0.5 * 10g * (0.4m/s)^2 + 0.5 * 30g * (-0.2m/s)^2 = 0.5 * 10g * v1^2 + 0.5 * 30g * v2^2
1.6 - 1.2 = 5v1^2 + 15v2^2
0.4 = 5v1^2 + 15v2^2
0.4 = 5v1^2 + 15v2^2
So we have the equations:
-2 = 10v1 + 30v2
0.4 = 5v1^2 + 15v2^2
Solving these two equations simultaneously, we will get v1 = -0.26 m/s and v2 = -0.07 m/s.
Therefore, the velocity of the smaller marble after the collision is 0.26 m/s to the left and the velocity of the larger marble after the collision is 0.07 m/s to the left.
(b) Change in momentum:
Change in momentum = (final momentum) - (initial momentum)
Change in momentum = (10g * -0.26 m/s + 30g * -0.07 m/s) - (10g * 0.4 m/s + 30g * -0.2 m/s)
Change in momentum = (-2.6g - 2.1g) - (4g - 6g)
Change in momentum = -4.7g - (-2g)
Change in momentum = -2.7g
The change in momentum is -2.7 g m/s.