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When the trolleys are released, it takes 0,3 s for the spring to unwind to its natural length - NSC Physical Sciences - Question 4 - 2016 - Paper 1

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When the trolleys are released, it takes 0,3 s for the spring to unwind to its natural length. Trolley Q then moves to the right at 4 m s⁻¹. 4.1 State the principle... show full transcript

Worked Solution & Example Answer:When the trolleys are released, it takes 0,3 s for the spring to unwind to its natural length - NSC Physical Sciences - Question 4 - 2016 - Paper 1

Step 1

State the principle of conservation of linear momentum in words.

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Answer

The total linear momentum in a closed system remains constant before and after an event, such as a collision.

Step 2

Velocity of trolley P after the trolleys are released.

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Answer

Using the conservation of momentum, we have:

mPvPi+mQvQi=mPvPf+mQvQfm_P v_{P_i} + m_Q v_{Q_i} = m_P v_{P_f} + m_Q v_{Q_f}

Substituting in the known values:

0=(0.4)(vPf)+2.4 0.4vPf=2.4 vPf=6 0 = (0.4)(v_{P_f}) + 2.4 \ 0.4v_{P_f} = -2.4 \ v_{P_f} = -6 \

Thus, the velocity of trolley P after release is 6extms16 ext{ m s}^{-1} to the left.

Step 3

Magnitude of the average force exerted by the spring on trolley Q.

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Answer

Using the formula for force:

F=maF = m a

We first need to find the acceleration of trolley Q:

Given the initial velocity vQi=0v_{Q_i} = 0 and final velocity vQf=4v_{Q_f} = 4 m/s, over rianglet=0.3 riangle t = 0.3 s:

a=vQfvQiΔt=400.3=13.33 m/s2a = \frac{v_{Q_f} - v_{Q_i}}{\Delta t} = \frac{4 - 0}{0.3} = 13.33 \text{ m/s}^2

Now substituting into the force formula:

F=(0.6)(13.33)=8extNF = (0.6)(13.33) = 8 ext{ N}

Step 4

Is this an elastic collision? Only answer YES or NO.

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Answer

NO

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