Figure 5 shows a simplified catapult used to hurl projectiles a long way - AQA - A-Level Physics - Question 4 - 2019 - Paper 1

To find the tension in the rope, we calculate the moments caused by the counterweight and the projectile. The moment due to the stone is:

$ext{Clockwise moment} = 610 ext{ kg} imes 9.81 ext{ m/s}^2 imes 1.5 ext{ m} = 8976 ext{ Nm}$

And the moment due to the projectile is:

$ext{Anticlockwise moment} = 250 ext{ N} imes 4.0 ext{ m} = 1000 ext{ Nm}$

Setting the moments equal for static equilibrium, we can find the tension using the relation:

$T = \frac{(8976 - 1000) ext{ Nm}}{1.5 ext{ m}} = 5310 ext{ N}$

To calculate the range of the projectile, we first need to determine the time of flight. Using the height of release:

$h = 7.5 ext{ m}$

Using the equation of motion for vertical drop:

$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 7.5}{9.81}} \approx 1.23 ext{ s}$

Next, we calculate the horizontal distance (range) traveled during this time at a velocity of 18 m/s:

$\text{Range} = v \cdot t = 18 ext{ m/s} \times 1.23 ext{ s} \approx 22.14 ext{ m}$

If the projectile is released just before the wooden beam is vertical, it will have less time to travel horizontally due to an immediate downward acceleration rather than from a horizontal launch.

1. Range may be smaller: The time spent in the air will decrease as the projectile will not have the same horizontal velocity component as when released horizontally.
2. Counterweight dynamics: The counterweight will fall slower, allowing less height for the projectile, hence reducing distance.
3. Conclusion: Overall, the range will be less than if released horizontally due to the decrease in horizontal velocity at the moment of release.