HELP A 0.060-kg tennis ball, moving with a speed of 5.38m/s , has a head-on coll

Question

HELP

A 0.060-kg tennis ball, moving with a speed of 5.38m/s , has a head-on collision with a 0.090-kg ball initially moving in the same direction at a speed of 3.50m/s . Assume that the collision is perfectly elastic.

1.A 0.060-kg tennis ball, moving with a speed of 5.38m/s ,has a head-on collision with a 0.090-kg ball initially moving in the same direction at a speed of 3.50m/s . Assume that the collision is perfectly elastic.

1.Determine the speed of the 0.060-

kg ball after the collision.

2.Determine the direction of the velocity of the 0.060-

kg ball after the collision.

3.Determine the speed of the 0.090-

kg ball after the collision.

4.Determine the direction of the velocity of the 0.090-

kg ball after the collision.

Explanation / Answer

elastic means momentum and kinetic energy are conserved

m1u1 + m2u2 = m1v1 +m2v2

0.06 x 4.38 + 0.09 x 3.5 = 0.06 x v1 + 0.09 x v2

1/2 *m1 * u1^2 + 1/2 * m2 * u2^2 = 1/2 *m1 * v1^2 + 1/2 * m2 * v2^2

0.5 * 0.06 * 4.38^2 + 0.5 * 0.09 * 3.5^2 = 0.5 * 0.06 * v1^2 + 0.5 * 0.09 * v2^2

simultaneous equations:

0.5778 = 0.06 v1 + 0.09 v2 (1)
2.2535 = 0.06v1^2 + 0.09 v2^2 (2)

(2) divided by (1)

2.025 / 0.54 = v1 + v2 = 3.90 (3)

so v2 = 3.90 - v1 by rearranging, and sub into (1)

0.5778 = 0.06 v1 + 0.09 * (3.90 - v1) which gives

0.5778 = 0.06v1 + 0.3510 - 0.09 v1 which gives

0.2268 = - 0.03 v1, so v1 = -7.56 m/s (so in opposite direction)

sub v1 = -7.56 into (1) to get v2 = 11.46 m/s in same direction as before

v2= 11.46m/s in same direction

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