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1 Calculate the speed of trolley Y. 4.2 State the principle of conservation of linear momentum. 4.3 Calculate the speed of trolley X before the collision.

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3 parts
4.1

2.5 m/s\text{2.5 m/s}

4.3

3.48 m/s\text{3.48 m/s}

4.4

42 N\text{42 N}

Step-by-step solution

9 steps

Part 4.1· 3 steps

  1. Identify the given values.

    Mass of trolley Y, mY=12 kgm_Y = 12 \, \text{kg}
    Kinetic energy of trolley Y, KEY=37.5 JKE_Y = 37.5 \, \text{J}

  2. Use the formula for kinetic energy to calculate the speed of trolley Y.

    KEY=12mYvY2KE_Y = \frac{1}{2}m_Y v_Y^2

    Substitute the known values:

    37.5 J=12(12 kg)vY237.5 \, \text{J} = \frac{1}{2} (12 \, \text{kg}) v_Y^2 37.5=6vY237.5 = 6 v_Y^2

  3. Solve for vYv_Y.

    vY2=37.56v_Y^2 = \frac{37.5}{6} vY2=6.25 m2/s2v_Y^2 = 6.25 \, \text{m}^2/\text{s}^2 vY=6.25 m2/s2v_Y = \sqrt{6.25 \, \text{m}^2/\text{s}^2} vY=2.5 m/sv_Y = 2.5 \, \text{m/s}

    The speed of trolley Y is 2.5 m/s\boxed{\text{2.5 m/s}}.

Part 4.2

The principle of conservation of linear momentum states that the total linear momentum of an isolated system remains constant in both magnitude and direction, provided no external forces act on the system.

Part 4.3

3.48 m/s\text{3.48 m/s}

  1. Identify the known values and the principle to be used.

    Mass of trolley X, mX=30 kgm_X = 30 \, \text{kg}
    Mass of trolley Y, mY=12 kgm_Y = 12 \, \text{kg}
    Speed of trolley Y before collision, vY=2.5 m/sv_Y = 2.5 \, \text{m/s} (from 4.1)
    Combined speed of the trolleys after collision, vf=3.2 m/sv_f = 3.2 \, \text{m/s}
    We need to find the speed of trolley X before collision, vXv_X.
    The collision is inelastic (trolleys stick together), so we use the conservation of linear momentum. Assume the direction to the right is positive.

  2. Apply the principle of conservation of linear momentum.

    Total momentum before collision = Total momentum after collision

    mXvX+mYvY=(mX+mY)vfm_X v_X + m_Y v_Y = (m_X + m_Y) v_f

  3. Substitute the values and solve for vXv_X.

    (30 kg)vX+(12 kg)(2.5 m/s)=(30 kg+12 kg)(3.2 m/s)(30 \, \text{kg}) v_X + (12 \, \text{kg}) (2.5 \, \text{m/s}) = (30 \, \text{kg} + 12 \, \text{kg}) (3.2 \, \text{m/s}) 30vX+30 kg⋅m/s=(42 kg)(3.2 m/s)30 v_X + 30 \, \text{kg}\cdot\text{m/s} = (42 \, \text{kg}) (3.2 \, \text{m/s}) 30vX+30=134.430 v_X + 30 = 134.4 30vX=134.4−3030 v_X = 134.4 - 30 30vX=104.430 v_X = 104.4 vX=104.430v_X = \frac{104.4}{30} vX=3.48 m/sv_X = 3.48 \, \text{m/s}

    The speed of trolley X before the collision is 3.48 m/s\boxed{\text{3.48 m/s}}.

Part 4.4

42 N\text{42 N}

  1. Identify the known values and the principle to be used.

    To calculate the force exerted by trolley X on trolley Y, we can use the impulse-momentum theorem for trolley Y.
    Mass of trolley Y, mY=12 kgm_Y = 12 \, \text{kg}
    Initial speed of trolley Y, vY=2.5 m/sv_Y = 2.5 \, \text{m/s} (from 4.1)
    Final speed of trolley Y (after collision), vf=3.2 m/sv_f = 3.2 \, \text{m/s}
    Collision time, Δt=0.2 s\Delta t = 0.2 \, \text{s}

  2. Apply the impulse-momentum theorem.

    The impulse exerted on trolley Y is equal to its change in momentum.

    FΔt=ΔpYF \Delta t = \Delta p_Y FΔt=mYvf−mYvYF \Delta t = m_Y v_f - m_Y v_Y F=mY(vf−vY)ΔtF = \frac{m_Y (v_f - v_Y)}{\Delta t}

  3. Substitute the values and solve for FF.

    F=12 kg(3.2 m/s−2.5 m/s)0.2 sF = \frac{12 \, \text{kg} (3.2 \, \text{m/s} - 2.5 \, \text{m/s})}{0.2 \, \text{s}} F=12 kg(0.7 m/s)0.2 sF = \frac{12 \, \text{kg} (0.7 \, \text{m/s})}{0.2 \, \text{s}} F=8.4 kg⋅m/s0.2 sF = \frac{8.4 \, \text{kg}\cdot\text{m/s}}{0.2 \, \text{s}} F=42 NF = 42 \, \text{N}

    The magnitude of the force that trolley X exerts on trolley Y is 42 N\boxed{\text{42 N}}.

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