Will give a medal to whoever can help!!

A projectile of mass m = 0.004 kg travelling at velocity v = 294 m/s impacts with a stationary mass M = 10 kg, that is suspended at the end of a massless cable of length l = 1.9 m. Immediately after impact, the projectile becomes embedded in the mass and the two objects can be treated as one composite of mass m+M. Calculate the maximum angle q (in degrees) the cable deviates from the vertical due to the impact of the projectile and the stationary mass. Note that q = 0 when mass M is suspended before impact. Make sure to give your answer in degrees.

Question

Will give a medal to whoever can help!!

A projectile of mass m = 0.004 kg travelling at velocity v = 294 m/s impacts with a stationary mass M = 10 kg, that is suspended at the end of a massless cable of length l = 1.9 m. Immediately after impact, the projectile becomes embedded in the mass and the two objects can be treated as one composite of mass m+M. Calculate the maximum angle q (in degrees) the cable deviates from the vertical due to the impact of the projectile and the stationary mass. Note that q = 0 when mass M is suspended before impact. Make sure to give your answer in degrees.

paxpolaris · Answer

just a guess of how you could solve it: 

kinetic energy before impact = potential energy at max angle

anonymous · Answer

how do you solve for the kinetic energy before impact?

paxpolaris · Answer

$$\frac 12 mv^2=\frac12\times 0.004 (294)^2 Joules$$

paxpolaris · Answer

it is the potential energy i am not sure of... not good with rotational physics :(

paxpolaris · Answer

|dw:1410556814311:dw|
$$\frac12\times 0.004 (294)^2 =(M+m)g\cdot1.9(1-\cos(q))$$

paxpolaris · Answer

i am getting 85.9 degrees

anonymous · Answer

thank you very much for your time and effort! :)