A bug is hovering in the air when a vehicle moving at 20m/s hits it. The colision is perfectly elastic, what is the speed of the bug after collision

Question

anonymous · Answer

Perfectly elastic collision = No change in kinetic energy (conservation of Kinetic energy?)

anonymous · Answer

$$KE_i=KE_f$$

anonymous · Answer

for an elastic collision with a stationary target
$$m _{1} V _{1i}^{} =  m _{1} *V _{1f}^{} +m _{2} *V _{2f}^{} $$
use:
$$V _{2f}^{} = \frac{ 2m _{1} }{ m _{1} +m _{2}}*V _{1i}^{} $$

anonymous · Answer

$$\frac{1}{2}m_cv_c^2=\frac{1}{2}m_bv_b'^2+\frac{1}{2}m_cv_c'^2 $$
$$m_cv_c=m_bv_b'+m_cv_c'$$

anonymous · Answer

$$\frac{m_cv_c-m_cv'_c}{m_b}=v'_b $$
$$\frac{2m_c}{m_c+m_b}*v_b=v_b'$$

anonymous · Answer

@Algebraic!  whered you come up with 2m at the top

anonymous · Answer

You equation itself can't be correct because i'm not given any masses so i can't solve it... somewhere mass must cancel

anonymous · Answer

k.

anonymous · Answer

m2~0

now try it

anonymous · Answer

going to need two simultaneous equations 
1) Kinetic energies will equal the same which i did now

anonymous · Answer

that makes sense but how'd you come to that conclusion

anonymous · Answer

it comes from using KEi =KEf to simplify m1v1i = m1v1f +m2v2f

anonymous · Answer

also you give no explaination of how you came to that equation.... .was it from simultaneously solving both the kinetic and the momentum equations?

anonymous · Answer

derivation is probably shown in your text.  it's just algebra. I've already done it for people several times on this site and it's invariably a waste of time, because they always say "oh, that was in my book actually" after you've gone through the steps for them.