A 0.311 kg tennis racket moving 30.3 m/s east makes an elastic collision with a 0.0570 kg ball moving 19.2 m/s east find the velocity of the tennis ball after the collision

The velocity of the ball after the collision is 64.5 m/s.

The west direction is regarded as the negative direction and the east direction as the positive direction.

Based on the given information,

•The mass of the tennis racket, m1 is 0.311 kg.

•The mass of the tennis ball, m2 is 0.057 kg.

•The initial velocity of the tennis racket, v1 is 30.3 m/s.

•The initial velocity of the tennis ball, v2 = -19.2 m/s.

•Let v3 be the velocity of the tennis racket after the collision and v4 be the velocity of the tennis ball after the collision.  

As the collision is elastic, thus, the coefficient of restitution, that is, e is 1.

According to the conservation of linear momentum,

Now putting the values we get,

Thus, the velocity of the tennis ball post collision is 64.5 m/s.

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38.0 m/s east

Explanation:

Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.311) (30.3) + (0.0570) (19.2) = (0.311) v₁ + (0.0570) v₂

10.52 = 0.311 v₁ + 0.0570 v₂

In an elastic collision, kinetic energy is conserved.

½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂²

m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²

(0.311) (30.3)² + (0.0570) (19.2)² = (0.311) v₁² + (0.0570) v₂²

306.5 = 0.311 v₁² + 0.0570 v₂²

Solve the system of equations.

0.311 v₁ = 10.52 − 0.0570 v₂

v₁ = 33.82 − 0.1833 v₂

306.5 = 0.311 (33.82 − 0.1833 v₂)² + 0.0570 v₂²

306.5 = 0.311 (1144 − 12.40 v₂ + 0.03360 v₂²) + 0.0570 v₂²

306.5 = 355.7 − 3.856 v₂ + 0.01045 v₂² + 0.0570 v₂²

0 = 0.06745 v₂² − 3.856 v₂ + 49.16

Use quadratic formula.

v₂ = [ 3.856 ± √(14.87 − 13.26) ] / 0.1349

v₂ = 19.2 or 38.0

We know v₂ isn't 19.2 m/s, so v₂ = 38.0 m/s.

pretty sure it's false

490 n

f=125n a=2.5m/s2