Question

*NOTE* The answers in this question are

NOT realistic. Don't second guess yourself!
A pitcher claims he can throw a 0.151 kg
baseball with as much momentum as a 3.22 g bullet moving with a speed of 1.90312 x
10° m/s.
What must be its speed if the pitcher's claim is valid?
Answer in units of m/s. Answer in units of ms.
004 (part 2 of 3) 1.0 points
What is the kinetic energy of the bullet?
Answer in units of J. Answer in units of J.
005 (part 3 of 3) 1.0 points
What is the kinetic energy of the ball?
Answer in units of J. Answer in units of J

Answer 1

Answer: It is not clear what is the power of 10 in the speed of the bullet but here is a general procedure for the question.

Explanation:

`P_(bullet)=(3.22*10^(-3)kg)*(10^(?)m/s)\\P_(baseball)=P_(bullet)\\`

Therefore we can write:

`(0.151*10^(-3) kg)*v_(baseball)=(3.22*10^(-3)kg)*(10^(?)m/s)`

From this using the relationship:

`v_(baseball)=((3.22*10^(-3)kg)*(10^?m/s))/(0.151*10^(-3)kg)`

I cannot give a precise answer since power of then for speed of the bullet is not clear. But using the above equation you can reach the answer. General form of the equation is therefore:

`v_(baseball)=(P_(bullet))/(m_(baseball))`

Where m is the mass of the ball which is 0.151 g.