A ball with a mass of 0.575kg is initially at rest. It is struck by a second bal

Question

A ball with a mass of 0.575kg is initially at rest. It is struck by a second ball having a mass of 0.430kg , initially moving with a velocity of 0.280m/s toward the right along the x axis. After the collision, the 0.430kg ball has a velocity of 0.215m/s at an angle of 36.3 degreea above the x axis in the first quadrant. Both balls move on a frictionless, horizontal surface.

1. What is the magnitude of the velocity of the 0.575kg ball after the collision? v=?m/s

2. What is the direction of the velocity of the 0.575kg ball after the collision? theta=? degrees below the x axis in the 4th quadrant

3. What is the change in the total kinetic energy of the two balls as a result of the collision?

Explanation / Answer

2dimensional elastic collisions

along x axis

m1*u1x + m2*u2x = m1*v1x + m2*v2x

(0.575*0) + (0.43*0.28) = (0.575*v1x) + (0.43*0.215*cos36.3)

0.1204 = 0.0745 + (0.575*v1x

v1x = 0.0798 m/s

along y axis

m1*u1y + m2*u2y = m1*v1y + m2*v2y

(0.575*0) + (0.43*0) = (0.575*v1y) + (0.43*0.215*sin36.3)

v1y = -0.055 m/s

1) v1 = sqrt(v1x^2 + v1y^2) = 0.0969 m/s

2) direction = tan^-1(v1y/v1x) = 34.57

3) K1 = 0.5*m2*u2^2 = 0.5*0.43*0.215*0.215 = 0.009938375 J

K2 = 0.5*m1*v1^2 + 0.5*m2*v2^2

K2 = (0.5*0.575*0.0969*0.0969) + (0.5*0.43*0.215*0.215)

K2 = 0.012637 J

dK = K2 - K1 = 0.002698625 J

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