A ball with a mass of 0.630 kg is initially at rest. It is struck by a second ba

Question

A ball with a mass of 0.630 kg is initially at rest. It is struck by a second ball having a mass of 0.400 kg , initially moving with a velocity of 0.280 m/s toward the right along the x axis. After the collision, the 0.400 kg ball has a velocity of 0.180 m/s at an angle of 37.1 above the x  axis in the first quadrant. Both balls move on a frictionless, horizontal surface.

Part A What is the magnitude of the velocity of the 0.630 kg ball after the collision? m/s Submit My Answers Give Up Part B What is the direction of the velocity of the 0.630 kg ball after the collision? o below the axis in the forth quadrant. Submit My Answers Give Up Part C What is the change in the total kinetic energy of the two balls as a result of the collision? Submit My Answers Give Up

Explanation / Answer

Use conservation of momentum
Momentum before Pb = Momentum after Pa
Pb = 0.400*0.280 = 0.112kgm/s
This is along the postive x axis meaning Pa can only have momentum along the positive x axis. That means that any vertical components must sum to zero for Pa=Pb
Let the velocity of the 0.630kg ball after the collision = V
and let it's direction or angle = x
We know that the horizontal component of the momentum of the 0.630kg ball must provide the missing horizontal momentum:
0.112 -0.400*0.180*cos37.1 = 0.630*V*cosx = 0.05457
V*cosx = 0.08661
We know 0.400*0.180*sin37.1 must be the canceled by vertical component of the momentum of the 0.630kg ball:
0.630*V*sinx = -0.400*0.180*sin37.1=> V*sinx = -0.400*0.180*sin37.1/0.630 = -0.06893
V*sinx/V*cosx = tanx => x = -38.51 degrees
V = 0.112 <---------------- A)
at -38.51degrees <----------- B)
So that makes sense: the smaller ball bounces above the x axis while the second ball travels below the x axis resulting in the sum of the vertical components of their momentum canceling.
Both balls have positive horizontal components of momentum that sum to the original horizontal momentum before the collision.
KEb = 0.5*0.400*0.280^2 = 0.01568J
KEa = 0.5*0.400*0.180^2 + 0.5*0.630*0.112^2 = 0.010431J
0.01568J - 0.01043J = 0.00525J <------- C)

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