A trolley of mass 1,5 kg is held stationary at point A at the top of a frictionless track. When the 1,5 kg trolley is released, it moves down the track. It passes point P at the bottom of the incline and collides with a stationary 2 kg trolley at point B. Refer to the diagram below. Ignore air resistance and rotational effects. 4.1 Use the principle of conservation of mechanical energy to calculate the speed of the 1,5 kg trolley at point P. When the two trolleys collide, they stick together and continue moving with constant velocity. 4.2 The principle of conservation of linear momentum is given by the incomplete statement below. In an ... system, the ... linear momentum is conserved. Rewrite the complete statement and fill in the missing words or phrases. 4.3 Calculate the speed of the combined trolleys immediately after the collision. 4.4 Calculate the distance travelled by the combined trolleys in 3 s after the collision.

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A trolley of mass 1,5 kg is held stationary at point A at the top of a frictionless track - NSC Physical Sciences - Question 4 - 2018 - Paper 1

Question 4

A trolley of mass 1,5 kg is held stationary at point A at the top of a frictionless track. When the 1,5 kg trolley is released, it moves down the track. It passes po... show full transcript

Worked Solution & Example Answer:A trolley of mass 1,5 kg is held stationary at point A at the top of a frictionless track - NSC Physical Sciences - Question 4 - 2018 - Paper 1

Step 1

4.1 Use the principle of conservation of mechanical energy to calculate the speed of the 1,5 kg trolley at point P.

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Answer

To calculate the speed of the 1,5 kg trolley at point P, we can apply the principle of conservation of mechanical energy:

At point A, the potential energy (PE) of the trolley is given by: $PE_A = m imes g imes h = 1.5 imes 9.81 imes 2.0$ While at point P, all of this potential energy is converted into kinetic energy (KE): $KE_P = \frac{1}{2} m v^2$ Setting the potential energy at point A equal to the kinetic energy at point P: $mgh = \frac{1}{2} mv^2$

Cancelling mass (m) from both sides: $gh = \frac{1}{2} v^2$

Rearranging to solve for v, we have: $v = \sqrt{2gh} = \sqrt{2 imes 9.81 imes 2.0}$ This results in: v \approx 6.26 , \text{m/s}

Step 2

4.2 The principle of conservation of linear momentum is given by the incomplete statement below.

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Answer

The complete statement filling in the missing word is:

"In an isolated system, the total linear momentum is conserved."

Step 3

4.3 Calculate the speed of the combined trolleys immediately after the collision.

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Answer

Using the conservation of momentum before and after the collision:

Let: m1 = 1.5 kg (mass of the first trolley), m2 = 2.0 kg (mass of the second trolley), v1 = 6.26 m/s (initial speed of the first trolley), v2 = 0 m/s (initial speed of the second trolley).

Momentum before collision: $p_{initial} = m_1 v_1 + m_2 v_2 = 1.5 imes 6.26 + 2.0 imes 0 = 9.39 \, \text{kg m/s}$

After the collision, the two trolleys stick together: Let (v_f) be their final speed: $p_{final} = (m_1 + m_2) v_f = (1.5 + 2.0) v_f$

Setting initial momentum equal to final momentum: $9.39 = (1.5 + 2.0) v_f$

Solving for (v_f): $v_f = \frac{9.39}{3.5} \approx 2.68 \text{ m/s}$

Step 4

4.4 Calculate the distance travelled by the combined trolleys in 3 s after the collision.

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Answer

Using the formula for distance travelled:

$\Delta x = v_f t + \frac{1}{2} a t^2$

Since there is no acceleration following the inelastic collision (they move together at constant speed):

The initial speed after collision (v_f = 2.68 , \text{m/s})
The acceleration (a = 0)

Substituting 3 seconds for (t): $\Delta x = 2.68 imes 3 + \frac{1}{2} \times 0 imes 3^2$ $\Delta x = 8.04 \, \text{m}$

A trolley of mass 1,5 kg is held stationary at point A at the top of a frictionless track - NSC Physical Sciences - Question 4 - 2018 - Paper 1

Worked Solution & Example Answer:A trolley of mass 1,5 kg is held stationary at point A at the top of a frictionless track - NSC Physical Sciences - Question 4 - 2018 - Paper 1

4.1 Use the principle of conservation of mechanical energy to calculate the speed of the 1,5 kg trolley at point P.

4.2 The principle of conservation of linear momentum is given by the incomplete statement below.

4.3 Calculate the speed of the combined trolleys immediately after the collision.

4.4 Calculate the distance travelled by the combined trolleys in 3 s after the collision.

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