Question

A ball with a mass of 0.605kg is initially at rest. It is struck by a second ball having a mass of 0.430kg , initially moving with a velocity of 0.220m/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 36.0? above the x  axis in the first quadrant. Both balls move on a frictionless, horizontal surface.

What is the magnitude of the velocity of the 0.605kg ball after the collision?

What is the direction of the velocity of the 0.605kg ball after the collision?

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

Answer 1

apply the law of conservation of momentum in Horizontal direction as

m1u1 + m2u2 =   m1v1 + M2v2 cos theta

here we need v1,

and u1 = 0

so

(0.605 * 0) +(0.430* 0.22) = (0.605 * v1) + (0.430 * 0.215 * COS 36)

0.605 V1 = (0.0946 -0.0747)

v1x = 0.0328 m/s

using vertical momentum conservation as

0+0 = 0.605 Vy + (0.430 * 0.215* sin 36)

Vy = 0.0543/0.605

Vy = -0.0898 m/s

so Vnet^2 = Vx^2 + Vy^2

Vnet^2 = (0.0328)^2 +(0.0898)^2

Vnet^2 =0.09143

Vnet = 0.0956 m/s

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direction tna theta = Vy/Vx

tan theta = -0.0898/0.0328

tan theta = -2.73

theta = -69.88 deg

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total chang in KE = 0.5 m1v1^2 + 0.5 m2v2^2 - (0.5 m1u1^2 + 0.5 m2u2^2)

total final KE = (0.5 * 0.605* 0.0956 *0.0956) +(0.5 * 0.430 * 0.215*0.215 * 0.81*0.81)

total final KE = 0.00928 J

total intial KE = 0.5*0.605 * 0* 0 + ( 0.5* 0.430 * 0.22*0.22)

total inital KE = 0.010406 J

so

Change in KE = 0.00928 -0.010406   = -0.0011 Jolues