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

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Figure 5 shows a simplified catapult used to hurl projectiles a long way. The counterweight is a wooden box full of stones attached to one end of the beam. The proj... show full transcript

Worked Solution & Example Answer:Figure 5 shows a simplified catapult used to hurl projectiles a long way - AQA - A-Level Physics - Question 4 - 2019 - Paper 1

Step 1

Suggest how the pivot position achieves this.

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Answer

The pivot position is strategically located at the center of mass of the beam and the empty wooden box. This ensures that the moments about the pivot due to the weights of these components balance each other, resulting in no net torque and thus, no effect on the tension in the rope.

Step 2

Calculate the tension in the rope.

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Answer

To calculate the tension in the rope, we first determine the clockwise and counterclockwise moments about the pivot:

  • Clockwise moment due to the counterweight: extClockwisemoment=610extkgimes9.81extm/s2imes1.5extm=8971.5extNm ext{Clockwise moment} = 610 ext{ kg} imes 9.81 ext{ m/s}^2 imes 1.5 ext{ m} = 8971.5 ext{ Nm}

  • Counterclockwise moment due to the projectile: extCounterclockwisemoment=250extNimes4extm=1000extNm ext{Counterclockwise moment} = 250 ext{ N} imes 4 ext{ m} = 1000 ext{ Nm}

Setting the sum of moments equal gives: Times4extm=8971.5extNm1000extNmT imes 4 ext{ m} = 8971.5 ext{ Nm} - 1000 ext{ Nm}
T = rac{7971.5 ext{ Nm}}{4 ext{ m}} = 1992.875 ext{ N}

Thus, the tension in the rope is approximately 1992.88 N.

Step 3

Calculate the range.

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To calculate the range, we can use the projectile motion equations. The time tt to fall a height of 7.5 m can be calculated with: h = rac{1}{2} g t^2
where g=9.81extm/s2g = 9.81 ext{ m/s}^2

Rearranging gives:

ightarrow t \approx 1.23 ext{ s}$$ The horizontal distance travelled (range) is given by: $$ ext{Range} = v imes t$$ Where $v = 18 ext{ m/s}$. Thus, $$ ext{Range} = 18 ext{ m/s} imes 1.23 ext{ s} \approx 22.14 ext{ m}$$ Thus, the range is approximately **22.14 m**.

Step 4

Discuss the effect this change has on the range of the catapult.

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Answer

When the projectile is released just before the beam becomes vertical, the initial velocity is directed upwards rather than horizontally. This change generally results in a greater component of the projectile's initial velocity aiding its ascent and causing it to gain height. The projectile will spend more time in the air, potentially increasing the overall range.

However, as the vertical component of velocity increases, the horizontal component is reduced, which might affect the distance it travels before hitting the ground. Therefore, while the total time in the air might be greater, the effective horizontal range may not increase as significantly. In many cases, this could either increase or decrease the range, depending on the balance of initial vertical and horizontal velocities.

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