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 find the speed of the 1.5 kg trolley at point P using conservation of mechanical energy, we set the potential energy at point A equal to the kinetic energy at point P.

The potential energy at point A is given by: $E_p = mgh = (1.5 ext{ kg})(9.8 ext{ m/s}^2)(2 ext{ m}) = 29.4 ext{ J}$

The kinetic energy at point P is: $E_k = \frac{1}{2}mv^2$

Setting these equal: $29.4 = \frac{1}{2}(1.5)v^2$

Solving for v: $v^2 = \frac{29.4 imes 2}{1.5} o v^2 = 39.2 o v = 6.26 ext{ m/s}$ Therefore, the speed of the trolley at point P is approximately 6.26 m/s.

Step 2

4.2 Rewrite the complete statement and fill in the missing words or phrases.

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Answer

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 linear momentum, we have: $m_{1}v_{1i} + m_{2}v_{2i} = (m_{1} + m_{2})v_{f}$

Where:

$m_{1} = 1.5 ext{ kg}$ (mass of trolley 1)
$v_{1i} = 6.26 ext{ m/s}$ (initial speed of trolley 1)
$m_{2} = 2 ext{ kg}$ (mass of trolley 2)
$v_{2i} = 0 ext{ m/s}$ (initial speed of trolley 2)

Substituting in: $(1.5)(6.26) + (2)(0) = (1.5 + 2)v_{f}$ $(9.39) = (3.5)v_{f}$

Solving for $v_{f}$ : $v_{f} = \frac{9.39}{3.5} = 2.68 ext{ m/s}$ Thus, the speed of the combined trolleys immediately after the collision is 2.68 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

To calculate the distance travelled by the combined trolleys in 3 seconds, we can use the formula: $\Delta x = v_{avg} \cdot t$

Where:

$v_{avg}$ is the average speed. Since the trolleys move with a constant speed after the collision: $v_{avg} = \frac{(2.68 + 2.68)}{2} = 2.68 ext{ m/s}$
$t = 3 ext{ s}$

Substituting in: $\Delta x = (2.68)(3) = 8.04 ext{ m}$ Therefore, the distance travelled by the combined trolleys in 3 seconds after the collision is 8.04 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 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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