Answer to Question #175913 in Physics for Omar

Question #175913

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



1
Expert's answer
2021-03-26T20:03:22-0400

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


PE=KE,PE=KE,12kx2=12mv2,\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2,v=kx2m=400 Nm(0.15 m)20.045 kg=14.14 ms.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:


m1v1i=m1v1f+m2v2f.m_1v_{1i}=m_1v_{1f}+m_2v_{2f}.

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


12m1v1i2=12m1v1f2+12m2v2f2.\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:


v2f=2m1v1i(m1+m2),v_{2f}=\dfrac{2m_1v_{1i}}{(m_1+m_2)},v2f=20.045 kg14.14 ms(0.045 kg+0.8 kg)=1.51 ms.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,PE=KE,mgh=12mv2f2,mgh=\dfrac{1}{2}mv_{2f}^2,h=v2f22g=(1.51 ms)229.8 ms2=0.116 m.h=\dfrac{v_{2f}^2}{2g}=\dfrac{(1.51\ \dfrac{m}{s})^2}{2\cdot9.8\ \dfrac{m}{s^2}}=0.116\ m.

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