Answer:
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.
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.
[tex]e = \frac{v4-v3}{v1-v2} \\1 = \frac{v4-v3}{30.3-(-19.2)} \\49.5 = v4-v3\\v3 = v4-49.5\\[/tex]
According to the conservation of linear momentum,
[tex]m1v1+m2v2=m1v3+m2v4[/tex]
Now putting the values we get,
[tex](0.311)(30.3)+(0.057)(-19.2) = (0.311)(v4-49.5)+(0.057)v4\\9.4233-1.0944=0.311v4-15.3945+0.057v4\\23.7234=0.368v4\\v4 = 64.5 m/s[/tex]
Thus, the velocity of the tennis ball post collision is 64.5 m/s.
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