Question

A 0.060-kg tennis ball, moving with a speed of5.62 m/s , has a head-on collision with a0.090-kg ball initially moving in the same direction at a speed of 3.06 m/s . Assume that the collision is perfectly elastic. Determine the speed of the 0.060-kg ball after the collision. Determine the direction of the velocity of the 0.060-kg ball after the collision. Determine the speed of the 0.090-kg ball after the collision.Determine the direction of the velocity of the 0.090-kg ball after the collision.

Answer #1

elastic means momentum and kinetic energy are conserved

m1u1 + m2u2 = m1v1 +m2v2

0.06 * 5.62 + 0.09 * 3.06 = 0.06 * v1 + 0.09 * v2

0.5 *m1 * u1^2 + 0.5 * m2 * u2^2 = 0.5 *m1 * v1^2 + 0.5 * m2 * v2^2

0.5 * 0.06 * 5.62^2 + 0.5 * 0.09 * 3.06^2 = 0.5 * 0.06 * v1^2 + 0.5 * 0.09 * v2^2

simultaneous equations:

0.6126 = 0.06 * v1 + 0.09 * v2 ........................................(1)

2.7378 = 0.06v1^2 + 0.09 v2^2 (2)

(2) divided by (1)

2.7378 / 0.6126 = v1 + v2 = 4.47 (3)

so v2 = 4.47 - v1 by rearranging, and sub into (1)

0.6126 = 0.06 v1 + 0.09 * (4.47 - v1) which gives

0.6126 = 0.06*v1 + 0.4023 - 0.09* v1 which gives

**0.2103 = - 0.03 * v1, so v1 = -7.01 m/s (so in opposite
direction)**

**sub v1 = -7.01 into (1) to get v2 = 11.48 m/s in same
direction as before**

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