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Two trolleys A and B of mass 3.2 kg and 2.6 kg respectively are held at rest on a straight horizontal, frictionless track, with a compressed spring between them, as shown in the diagram below - NSC Physical Sciences - Question 4 - 2024 - Paper 1

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Two trolleys A and B of mass 3.2 kg and 2.6 kg respectively are held at rest on a straight horizontal, frictionless track, with a compressed spring between them, as ... show full transcript

Worked Solution & Example Answer:Two trolleys A and B of mass 3.2 kg and 2.6 kg respectively are held at rest on a straight horizontal, frictionless track, with a compressed spring between them, as shown in the diagram below - NSC Physical Sciences - Question 4 - 2024 - Paper 1

Step 1

State the principle of conservation of linear momentum in words.

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Answer

In an isolated system, the total (linear) momentum remains constant. This means that the total momentum before the event is equal to the total momentum after the event.

Step 2

Calculate the distance travelled by trolley B in 1.3 s.

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Answer

To calculate the distance travelled by trolley B, we can use the formula:

extDistance=extVelocityimesextTime ext{Distance} = ext{Velocity} imes ext{Time}

Given that trolley A moves with a constant velocity of 0.4 m s^-1 to the left, we need to determine the velocity of trolley B. According to the information provided, the time taken for trolley B to reach the end of the track is 1.3 s, and we can express the distance for trolley B as:

extDistanceB=vBimest=vBimes1.3 ext{Distance}_B = v_B imes t= v_B imes 1.3

From the conservation of momentum, we have:

3.2imes(0.4)+2.6imesvB=03.2 imes (-0.4) + 2.6 imes v_B = 0

Solving this gives:

v_B = rac{3.2 imes 0.4}{2.6} = 0.4923 ext{ m s}^{-1}

Thus, the distance:

extDistanceB=0.4923imes1.3=0.64extm ext{Distance}_B = 0.4923 imes 1.3 = 0.64 ext{ m}

Step 3

Calculate the time it took the spring to extend to its natural length.

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Answer

The average force exerted by the spring on each trolley is given as 4.2 N. Using Newton's second law of motion, we can express the relationship as:

F=mimesaF = m imes a

For trolley A:

4.2 = 3.2 imes a_A \ \text{Thus, } a_A = rac{4.2}{3.2} = 1.3125 ext{ m s}^{-2}

Now, using the kinematic equations, we have:

v = u + at \ v = 0 ext{ (initial velocity)} + (1.3125 imes t)\ ext{Using the final velocity from previous calculations,}\ 0.49 = 0 + (1.3125 imes t)\ t = rac{0.49}{1.3125} = 0.37 ext{ s}

Step 4

How does the magnitude of the velocity of trolley B after the spring has fallen to the track? Write only GREATER THAN, LESS THAN or EQUAL TO. Explain the answer.

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Answer

The magnitude of the velocity of trolley B after the spring has fallen to the track will be LESS THAN the velocity of trolley A after the spring has extended because trolley C, which has a larger mass, is now present in the system. The larger mass will result in a smaller acceleration for trolley C after the same force acts upon it, thus reducing the final velocity of trolley B after the spring is released.

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