Posted by lib on .
Your physical therapist throws a baseball to you at a certain speed and you catch it. To increase the difficulty, the therapist is going to throw you a medicine ball whose mass is ten times the mass of the baseball. You are given the following choices. Rank these from easiest to hardest to catch.
1. The medicine ball thrown at the same speed as the baseball
2. The medicine ball thrown with the same momentum as the baseball.
3. The medicine ball thrown with the same kinetic energy as the baseball.

physics 
Eli,
The sequence would be 2,3,1. Looking at a specific variable, lets say the velocity of the larger mass.
For the first kind of catch, the velocity of the medicine ball is the same, so it should be the hardest as there is no compensation for difficulty due to the increase in mass.
The second kind of catch, is the easiest as the relationship between momentum is (m1)*(v1)=(m2)*(v2).
Since we know that the mass of the medicine ball is ten time the mass of the baseball, this can be rewritten as v2=(1/10)v1. (The 1/10 is from the mass relations, and the mass values cancel out (m1/10m1).
The third kind of catch is a little bit trickier to calculate. By substituting 10(m1) in for (m2) though, it can be derived. The equation for kinetic energy is KE=.5mv^2. If the KE is equal, then the statement (m1)(v1)^2=(m2)(v2)^2 should be true. Since 10m1=m2, it can be further derived that .1(v1)^2=(v2)^2. So V2=sqrt(.1*(v1)^2). This is greater than the second method(since sqrt(.1)>.1), but less than the first. So the order is 2, 3 then finally 1. 
physics 
Cyberian.pk,
Think about that answer. If you catch something 10 times heavier going at the same speed, it's kinetic energy and its momentum are both going to be multiplied by 10. So given that you have a choice between the same momentum, the same kinetic energy, or 10 times as much momentum and kinetic energy, I'd say the choice "a" is by far the worst option. In order to keep the same momentum and kinetic energy for the heavier ball, it's speed would have to be reduced, so b and c would make much more sense.