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A 50.0 kg child is riding in a 120.0 kg bumper car at an amusement park. The child drives the bumper car at 4.5 m/s head-on into the wall of the ride and rebounds at the speed. What is the magnitude of the change in momentum of the child in the bumper car?

Sagot :

Answer:

Explanation:

Total mass of child and car = 50 + 120 = 170 kg .

initial velocity u = 4.5 m/s

initial momentum = mass x velocity = 170 x 4.5 = 765 kg m/s

final velocity = - 4.5 m/s [ - ve is due to reversal of direction ]

final momentum = - 765 kg m/s

change in momentum = final momentum - initial momentum

= - 765 - ( 765 )

= - 1530 kg m/s .

Momentum of a body is the product of mass and the velocity of the body.

The magnitude of the change in the momentum of the child in the bumper car is 1530 kg-m/s.

What is change in momentum?

The change in momentum of a object is the difference of the final momentum to the initial momentum.

Given information-

The mass of the child is 50 kg.

The mass of the bumper car is 120 kg.

The initial velocity of the bumper car is 4.5 m/s.

Momentum of a body is the product of mass and the velocity of the body. The initial momentum of the child in the bumper car is,

[tex]p_i=(m_c+m_b)u[/tex]

Here, [tex]m_c[/tex] is the mass of child and [tex]m_b[/tex] is the mass of the bumper car. Also [tex]u[/tex] is the initial velocity,

[tex]p_i=(50+120)\times4.5\\p_i=765[/tex]

Thus the initial momentum of the child in the bumper car is 765 kg-m/s.

When the bumper car head-on into the wall of the ride then it rebounds at the speed. Thus the final momentum will be opposite in the direction. Thus final momentum is,

[tex]p_f=-p_i\\p_f=-765[/tex]

As the change in momentum of bumper car is the difference of the its final momentum to the initial momentum. thus,

[tex]\Delta p=pf-pi\\\Delta p=-765-765\\\Delta p=-1530[/tex]

Hence the magnitude of the change in the momentum of the child in the bumper car is 1530 kg-m/s.

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