Use one of these equations. Show all steps. p = mv; PE = mgh; KE = (1/2)mv2; P=W
ID: 249098 • Letter: U
Question
Use one of these equations. Show all steps. p = mv;
PE = mgh; KE = (1/2)mv2;
P=W/t; W=Fdcos(theta);
F = ( p/ t);x-xo = vot + 1/2 a t2 ;
v=vo+at ;
v2=vo2+2a(x-xo);
vavg=(v+vo)/2
A 50 kg woman skating to the right at 3 m/s skates into a a 20 kg boy travelling 2 m/s to the left. After they meet, they continue skating arm-in-arm on the frictionless ice. a) What is the speed and direction with which the pair moves off after the collision? b) Determine wheter or not the collision is elastic (Show your caluculations). c) Discuss whether or not your answers to parts a) and b) are reasonable. Include specific reasons why you think the answers are or are not reasonable.
Explanation / Answer
Here,
m1 = 50 kg, v1 = 3 m/s
m2 = 20 kg, v2 = -2 m/s (we are considering rightdirection as positive)
After meeting, both will move at the same speed v.
So, considering conservation of momentum,
50*3 + 20*(-2) = (50+20)v
=> 150-40 = 70v
=> v = 110/70 = 1.57 m/s
(a) So, the pair will move with a speed of 1.57 m/s in right direction.
(b) Kinetic energy of the two bodies before meeting = 0.5*50*3^2 + 0.5*20*2^2 = 225+40 = 265 J
Kinetic energy of the two bodies after meeting = 0.5*(50+20)*1.57^2 = 86.27 J
Since the kinetic energy before meeting is not equal to after meeting. Hence the collision is inelastic.
(c) This is self explanatory that the answers at (a) and (b) is reasonable.
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