Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

Anonymous asked in Science & MathematicsPhysics · 8 years ago

Physics question, please help, will really appreciate it :)?

2 boys, John and Peter have masses of 50kg and 80kg. They stand on a stationary trolley with a mass of 180kg. The trolley is free to move from left to right on a horizontal surface. They boys simultaneously jump in opposite directions from opposite ends of the trolley. Both boys have an initial relative speed of 3 m.s-1 to the ground.

John Peter

Mass: 50kg Mass: 80kg

Speed prior to jump: 0 m.s-1 Speed prior to jump: 0 m.s-1

Speed after jump: 3 m.s-1 Speed after jump: 3 m.s-1

1) Calculate the magnitude and direction of the velocity with which the trolley starts to move immediately after the boys jumped off the trolley.

2) Give a reason why the velocity calculated in Question 1 does not remain constant after the boys jumped off.

3) Use Newton's Second Law and explain why the trolley moves in the direction determined in Question 1.

4) While peter jumps off the trolley, he pushes down on the trolley for 0,2s. Calculate the magnitude and direction of the force that Peter exerts on the trolley during this time.

5) Explain why Peter would accelerate to the right if the force that he exerts on the trolley has the same magnitude, but is exerted in the opposite direction.

Really hope you can help with this :)

1 Answer

  • ?
    Lv 7
    8 years ago
    Favorite Answer

    Let's say John is on the left and jumps left and Peter is on the right and jumps right

    Conservation of momentum

    Let John be m1

    Let Peter be m2

    Let trolley be m3

    Let right be the positive direction

    m1u1 + m2u2 + m3u3 = m1v1 + m2v2 + m3v3

    m1(0)+ m2(0) + m3(0) = m1v1 + m2v2 + m3v3

    0 = m1v1 + m2v2 + m3v3

    v3 = [- m1v1 - m2v2] / m3

    v3 = [- 50(-3) - 80(3)] / 180

    v3 = [150 - 240] / 180

    v3 = -0.5 m/s

    v3 = 0.5 m/s towards John ===>ANS(1)


    The velocity does not remain constant for one of two reasons.

    Either John sprained his ankle while landing and was unable to get out of the trolley's path. The trolley rolled over his legs causing compound fractures of both femurs. The energy of breaking bone and crushing flesh came from a decrease in kinetic energy of the trolley.


    Friction in the trolley axles slowed it down by converting kinetic energy to thermal energy in the system.


    As Peter has a larger mass than John, his mass requires a larger force to accelerate to a given velocity in a given time period.

    From Newton's first law and the viewpoint of the trolley, the trolley must supply a larger force to Peter than to John.

    From Newton's second law, the larger force on Peter creates an unbalance of forces on the trolley with the larger force being in John's direction. The trolley accelerates in the direction of the larger force.


    An impulse equals a change in momentum

    FΔt = mΔv

    F = mΔv / Δt

    Actually, pushing down on the trolley will not cause it to move. We must make the logic jump that a downward push also allows the horizontal friction force to increase to the necessary level to accelerate Peter to 3 m/s relative to the ground.

    as time starts at the beginning of the leap and initial velocities are zero, we can use the mass, time and maximum velocity of Peter to find the force.

    F = mv / t

    F = 80(3) / 0.2

    F = 1200 N


    Again, the combination of Newton's First and Second laws combine to get Peter moving to the right for a leftward movement of the trolley.

Still have questions? Get your answers by asking now.