Question

a toy gun fires a 0.045 kg projectile by using a compressed spring (k=400n/m). the spring is compressed 15 cm. the projectile bounces off a stationary mass of 0.80 kg in the head on elastic collision. The mass is suspended to the ceiling by a string which allows to swing to an unknown height. Determine the unknown height

Answer 1

Let's first find the initial velocity of the projectile from the law of conservation of energy:

$$PE=KE,$$ $$\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2,$$ $$v=\sqrt{\dfrac{kx^2}{m}}=\sqrt{\dfrac{400\ \dfrac{N}{m}\cdot(0.15\ m)^2}{0.045\ kg}}=14.14\ \dfrac{m}{s}.$$

Then, we can find the final velocity of the stationary mass after the collision from the law of conservation of momentum:

$$m_1v_{1i}=m_1v_{1f}+m_2v_{2f}.$$

Since collision is elastic, kinetic energy is conserved and we can write:

$$\dfrac{1}{2}m_1v_{1i}^2=\dfrac{1}{2}m_1v_{1f}^2+\dfrac{1}{2}m_2v_{2f}^2.$$

This formula gives us an additional relationship between velocities. Therefore, with the help of these two formulas we can find the final velocity of the stationary mass after the collision:

$$v_{2f}=\dfrac{2m_1v_{1i}}{(m_1+m_2)},$$ $$v_{2f}=\dfrac{2\cdot0.045\ kg\cdot14.14\ \dfrac{m}{s}}{(0.045\ kg+0.8\ kg)}=1.51\ \dfrac{m}{s}.$$

Finally, we can find the unknown height from the law of conservation of energy:

$$PE=KE,$$ $$mgh=\dfrac{1}{2}mv_{2f}^2,$$ $$h=\dfrac{v_{2f}^2}{2g}=\dfrac{(1.51\ \dfrac{m}{s})^2}{2\cdot9.8\ \dfrac{m}{s^2}}=0.116\ m.$$