A ball with a mass of 0.570 kg is initially at rest. It is struck by a second ba

Question

A ball with a mass of 0.570 kg is initially at rest. It is struck by a second ball having a mass of 0.405 kg , initially moving with a velocity of 0.275 m/s toward the right along the x axis. After the collision, the 0.405 kg ball has a velocity of 0.170 m/s at an angle of 35.9 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 0.570 kg ball after the collision? Part B What is the direction of the velocity of the 0.570 kg ball after the collision? Part C What is the change in the total kinetic energy of the two balls as a result of the collision?

Explanation / Answer

here,

mass , m1 = 0.57 kg

mass m2 = 0.405 kg at u2 = 0.275 m/s

after the collison,

v2 = 0.405 at theta2 = 35.9 degree

A)

let the velocity of ball 1 be v1 at angle theta1

using conservation of momentum along x axis

m2 * u2 = m1 * v1 * cos(35.9) + m2 * v2 * cos(theta2)

0.405 * 0.275 = 0.57 * v1 * cos(theta1) + 0.405 * 0.17 * cos(35.9) ....(1)

and

using conservation of momentum along y axis

m1*v1*sin(theta1) = m2 * v2 * sin(theta2)

0.57 * v1* sin(theta1) = 0.405 * 0.17 * sin(35.9) ....(2)

from equation (1) and (2)

v1 = 0.25 m/s at theta = 16.46 degree

A)

the magnitude of velocity is 0.25 m/s

B)

the direction of velocity after collsion is 16.46 degree clockwise from the axis of collison

C)

change in total kinetic energy , TE = 0.5 * ( m1*v1^2 + m2*v2^2 - m2*u2^2)

TE = 0.5 * ( 0.57*0.25^2 + 0.405 * 0.17^2 - 0.405* 0.275^2)

TE = 8.35 * 10^-3 J

the change in the total kinetic energy of the two balls as a result of the collision is 8.35 * 10^-3 J

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