Question

A catapult is made by connecting a light elastic cord of natural length 20 cm and force constant 100 N/m between two fixed supports, which are distance 20 cm apart. A stone of mass 1 kg is placed at the center of the cord, which is pulled back a distance 7.5 cm and then released from rest. Find the speed with which the stone is projected by the catapult.

• A. $\frac{1}{2\sqrt{2}}ms^{-1}$
• B. $\sqrt{2} ms^{-1}$
• C.
  0. 5 $ms^{-1}$
• D. $1ms^{-1}$

Answer 1

Answer: (C)

0.5 $ms^{-1}$

Answer 2

The natural length of the cord is $L_0 = 20$ cm. The distance between the supports is $D = 20$ cm. When the stone is pulled back by $x = 7.5$ cm, the length of each segment of the cord is $l = \sqrt{(D/2)^2 + x^2} = \sqrt{(10 \text{ cm})^2 + (7.5 \text{ cm})^2} = 12.5$ cm. The total stretched length is $L_{stretched} = 2l = 25$ cm. The total extension is $\Delta L = L_{stretched} - L_0 = 25 \text{ cm} - 20 \text{ cm} = 5$ cm $= 0.05$ m. The potential energy stored in the cord is $PE = \frac{1}{2} k (\Delta L)^2 = \frac{1}{2} (100 \text{ N/m}) (0.05 \text{ m})^2 = 0.125$ J. This potential energy is converted into kinetic energy ($KE = \frac{1}{2} m v^2$) of the stone when it reaches the equilibrium position. Equating $PE = KE$, we get $0.125 \text{ J} = \frac{1}{2} (1 \text{ kg}) v^2$. Solving for $v$, we get $v^2 = 0.25 \text{ m}^2/\text{s}^2$, so $v = 0.5$ m/s.