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A teacher demonstrates the principle of conservation of linear momentum using two trolleys - NSC Physical Sciences - Question 4 - 2017 - Paper 1

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A teacher demonstrates the principle of conservation of linear momentum using two trolleys. The teacher first places the trolleys, A and B, some distance apart on a ... show full transcript

Worked Solution & Example Answer:A teacher demonstrates the principle of conservation of linear momentum using two trolleys - NSC Physical Sciences - Question 4 - 2017 - Paper 1

Step 1

Use the table above to prove that trolley A is moving at constant velocity before it collides with trolley B.

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Answer

To prove that trolley A is moving at constant velocity, we can use the formula for velocity, which is calculated by the change in position divided by the change in time:

v=ΔxΔtv = \frac{\Delta x}{\Delta t}

From the table:

  • For the interval from 0.0 m to 0.2 m, the time taken is 0.4 s:

    • Velocity = ( \frac{0.2 - 0.0}{0.4 - 0.0} = \frac{0.2}{0.4} = 0.5 , m/s )

  • For the interval from 0.2 m to 0.4 m, the time taken is also 0.4 s:

    • Velocity = ( \frac{0.4 - 0.2}{0.4 - 0.0} = \frac{0.2}{0.4} = 0.5 , m/s )

  • For the interval from 0.4 m to 0.6 m:

    • Velocity = ( \frac{0.6 - 0.4}{0.4} = 0.5 , m/s )

Since all calculated velocities are equal, we can conclude that trolley A is moving at a constant velocity of 0.5 m/s.

Step 2

State the principle of conservation of linear momentum.

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Answer

The principle of conservation of linear momentum states that in a closed (isolated) system, the total linear momentum remains constant, provided no external forces act on it. This means that the total momentum before a collision is equal to the total momentum after the collision.

Step 3

Calculate the magnitude of the average net force exerted on trolley B by trolley A.

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Answer

To calculate the average net force exerted on trolley B by trolley A, we first need to determine the change in momentum during the collision:

Given:

  • Mass of trolley A: mA=3.5 kgm_A = 3.5 \, kg\
  • Initial velocity of trolley A before collision: viA=0.5 m/sv_{iA} = 0.5 \, m/s\
  • Final combined velocity after the collision: vf=0.184 m/sv_f = 0.184 \, m/s\
  • Collision time: t=0.5 st = 0.5 \, s\

The change in momentum is calculated as:

Δp=mA(vf−viA)\Delta p = m_A (v_f - v_{iA})

Substituting the given values:

Δp=3.5 kg(0.184−0.5)=3.5 kg(−0.316)=−1.106 kg m/s\Delta p = 3.5 \, kg (0.184 - 0.5) = 3.5 \, kg (-0.316) = -1.106 \, kg \, m/s

Now, the net force can be calculated using the impulse-momentum theorem:

Fnet=ΔpΔtF_{net} = \frac{\Delta p}{\Delta t}

Substituting the values:

Fnet=−1.1060.5=−2.212 NF_{net} = \frac{-1.106}{0.5} = -2.212 \, N

Thus, the magnitude of the average net force exerted on trolley B by trolley A is approximately 2.21 N2.21 \, N.

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