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A racquet ball with mass m = 0.256 kg is moving toward the wall at v = 11.8 m/s and at an angle of θ = 29° with respect to the horizontal. The ball makes a perfectly elastic collision with the solid, frictionless wall and rebounds at the same angle with respect to the horizontal. The ball is in contact with the wall for t = 0.066 s.1)What is the magnitude of the initial momentum of the racquet ball?kg-m/s2) What is the magnitude of the change in momentum of the racquet ball?kg-m/s3) What is the magnitude of the average force the wall exerts on the racquet ball?N4) ballhitswall2 Now the racquet ball is moving straight toward the wall at a velocity of vi = 11.8 m/s. The ball makes an inelastic collision with the solid wall and leaves the wall in the opposite direction at vf = -7.8 m/s. The ball exerts the same average force on the ball as before.What is the magnitude of the change in momentum of the racquet ball?kg-m/s5) What is the time the ball is in contact with the wall?s6). What is the change in kinetic energy of the racquet ball?J

Respuesta :

Answer:

Part a)

[tex]P = 5.72 kg m/s[/tex]

Part b)

[tex]\Delta P = 2.93 kg m/s[/tex]

Part c)

[tex]F = 44.4 N[/tex]

Part d)

[tex]\Delta P = 5.02 kg m/s[/tex]

Part e)

[tex]\Delta t = 0.113 s[/tex]

Part f)

[tex]\Delta K = 0[/tex]

Explanation:

As we know that initial velocity of the ball is given as

[tex]v = 11.8 cos29 \hat i + 11.8 sin29 \hat j[/tex]

[tex]v_i = 10.3 \hat i + 5.72 \hat j[/tex]

Now final velocity of the system is given as

[tex]v_f = 10.3\hat i - 5.72\hat j[/tex]

Part a)

now magnitude of initial momentum is given as

[tex]P = mv[/tex]

[tex]P = 0.256(11.8)[/tex]

[tex]P = 5.72 kg m/s[/tex]

Part b)

Change in momentum is given as

[tex]\Delta P = m(v_f - v_i)[/tex]

[tex]\Delta P = 0.256(5.72 + 5.72)[/tex]

[tex]\Delta P = 2.93 kg m/s[/tex]

Part c)

As we know that average force is defined as the rate of change in momentum

so here we have

[tex]F = \frac{\Delta P}{\Delta t}[/tex]

[tex]F = \frac{2.93}{0.066}[/tex]

[tex]F = 44.4 N[/tex]

Part d)

Magnitude of change in momentum is given as

[tex]\Delta P = m(v_f - v_i)[/tex]

[tex]\Delta P = 0.256(7.8 + 11.8)[/tex]

[tex]\Delta P = 5.02 kg m/s[/tex]

Part e)

As we know that in 2nd case the force is same as the initial force

so we will have

[tex]\frac{\Delta P}{\Delta t} = F[/tex]

[tex]\frac{5.02}{\Delta t} = 44.4[/tex]

[tex]\Delta t = 0.113 s[/tex]

Part f)

Since this is elastic collision so change in kinetic energy must be ZERO

[tex]\Delta K = 0[/tex]

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