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

To calculate the tension in the rope, we need to consider the balance of moments about the pivot. The clockwise moment due to the counterweight (stones) is equal to the anti-clockwise moment due to the weight of the projectile.

The total weight of the stones is calculated as:

$610 ext{ kg} imes 9.81 ext{ m/s}^2 = 5976.6 ext{ N}$

The anti-clockwise moment using the distance 4 m is:

$ext{Anti-clockwise moment} = 250 ext{ N} imes 4 ext{ m} = 1000 ext{ Nm}$

Setting clockwise moments equal to anti-clockwise moments:

$T imes 1.5 ext{ m} = 5976.6 ext{ Nm} - 1000 ext{ Nm}$

Therefore,

T = rac{4976.6 ext{ Nm}}{1.5 ext{ m}} = 3317.73 ext{ N}

Thus, the tension in the rope is approximately 3318 N.

To find the range of the projectile, we first calculate the time of flight using the height of 7.5 m.

Using the formula for vertical motion:

h = rac{1}{2} g t^2

where:

• $h = 7.5$ m,
• $g = 9.81$ m/s².[solve for t]

Rearranging gives:

t = ext{sqrt}rac{2h}{g} = ext{sqrt}rac{2 imes 7.5}{9.81} ext{ ans = 1.23 s}

Next, we calculate the horizontal distance covered:

$ext{range} = v imes t = 18 ext{ m/s} imes 1.23 ext{ s} = 22.14 ext{ m}$

So, the range is approximately 22.14 m.

When the projectile is released just before the beam is vertical, it will depart with a vertical component of velocity, rather than a horizontal one. As a result, this vertical component will reduce the overall initial horizontal velocity of the projectile.

Consequently, the range will be less compared to when the projectile is launched horizontally because:

• The time of flight may be shorter if the projectile descends vertically immediately after release.
• The horizontal distance traveled will be less due to the loss of horizontal momentum.

Overall, the change in the release position leads to a decreased range for the projectile.