A 66.2 kg ice skater moving to the right with a velocity of 2.97 m/s throws a 0.150 kg snowball to the right with a velocity of 32.4 m/s relative to the ground. What is the velocity of the ice skater after throwing the snowballWhat is the velocity of the second skater after catching the snowball in a perfectly inelastic collision? Answer in units of m/s

Answer

d2= 2.89m/s

V2 =0.3/(H+0.15)m/s

Explanation:

Let M mass of ice skatter

N mass of snow ball ,and d1 velocity of ice skatter before throwing and g1 velocity of snowball before been thrown,d2 velocity after throwing and g2 after thrown

Hence M=66.2kg,N=0.15kg,d1=2.97m/sd2=?,g1=2.97m/s and g2=32.4m/s

Recall that momentum p=mass m*velocity v writing for the symbol above

M*d1 + N*g1=M*d2+M*g2

66.2(kg)*2.97m/s+0.15kg*2.97m/s=66.2(kg)*d2+0.15kg*32.4m/s

196.614kgm/s+0.45kgm/s=66.2d2+4.86kgm/s

196.45kgm/s-4.86=66.2d2

191.56=66.2d2

d2=2.89m/s

Part 2

Since the mass of the second skatter was not given let's assume it as Hkg now

Mass ice skatter 2 is Hkg

Iniatial velocity is 0

N is 0.15kg,g2 is 32.4m/s

V2 second skatter velocity after catching the snow

Also

H*0+N*g2=(H+N)V2 perfectly inelastic condition

H(kg)*0+.15kg*32.4m/s=(H+0.15)V2

0.3/(H+0.15)m/s=V2

Hence V2 can be found when mass of second skatter is specified with the relationship above

i am pretty sure it is c because when people are closer together, they cannot move as freely. picture being in a mostly full elevator, for instance. the people are right next to you, and you can barely move.