Question

The moon’s orbital speed around Earth is 3.680 × 10^3 km/h. Suppose the moon suffers a perfectly elastic collision with a comet whose mass is 50.0 percent that of the moon. (A partially inelastic collision would be a much more realistic event.) After the collision, the moon moves with a speed of−4.40 × 10^2 km/h, while the comet moves away from the moon at−5.740 × 10^3 km/h. What is the comet’s speed before the collision?

Answer 1

Speed of comet before collision is

`v_{2_(i)}=-2.5*10^(3)\quad km/h`

Step-by-step explanation:

Correction: (As stated after collision comet moves away from moon so velocity of moon and moon and comet must be opposite in direction. as spped of moon after collision is −4.40 × 10^2km/h so that comet's must be 5.740 × 10^3km/h instead of -5.740 × 10^3km/h)

Solution:

`mass \quad of\quad moon = m_(1)\\\\mass\quad of \quad comet = m_(2) = 0.5 m_(1)\\\\speed\quad of\quad moon\quad before\quad collision = v_{1_(i)}=3.680* 10^3 km/h\\\\speed \quad of\quad moon\quad after\quad collision=v_{1_(f)} = -4.40 * 10^2 km/h\\\\speed\quad of\quad comet\quad after\quad collision =v_{2_(f)} =5.740 * 10^3 km/h`

Case is considered as partially inelastic collision, by conservation of momentum

`m_(1)v_{1_(i)}+m_(2)v_{2_(i)}=m_(1)v_{1_(f)}+m_(2)v_{2_(f)}\\\\m_(1)v_{1_(i)}+0.5m_(1)v_{2_(i)}=m_(1)v_{1_(f)}+0.5m_(1)v_{2_(f)}\\\\v_{1_(i)}+0.5v_{2_(i)}=v_{1_(f)}+0.5v_{2_(f)}\\\\v_{2_(i)}=2(v_{1_(f)}+0.5v_{2_(f)}-v_{1_(i)})\\\\v_{2_(i)}=2(-4.40 * 10^2+0.5(5.740 * 10^3)-3.680 * 10^3 )\\\\v_{2_(i)}=-2.5*10^(3)\quad km/h`