Physics Question! Help!
A pitcher claims he can throw a 0.187 kg baseball with as much momentum as a 3.12 g bullet moving with a speed of 1.11436×105 m/s.
1. What must be its speed if the pitcher’s claim is valid? Answer in units of m/s.
2. What is the kinetic energy of the bullet? Answer in units of J
3. What is the kinetic energy of the ball? Answer in units of J
1 Answer | Add Yours
Alright, the formula for momentum is p = mv, where the "p" is momentum, the "m" is mass, and the "v" is velocity (speed with direction). So, to find the velocity of the ball, let's first calculate the momentum of the bullet:
3.12g x 1.11436 x 105 m/s = 3.4768 x 105 J or 347, 680 J
So, to calculate the velocity of the ball, simply divide the momentum of the bullet by the mass of the ball:
347,680/187 = 1859.25 m/s
Now, that's what I call a FAST BALL! One thousand, eight hundred and fifty nine meters per second! Sign him up!
The formula for kinetic energy is KE = 1/2mv2, where the KE is kinetic energy, the m again is mass, the v is velocity, so:
1/2 (3.12)(111,436)2 = 19,372,052,069.7 J, for the bullet.
The kinetic energy of the ball, provided the pitcher can throw it that fast, would be the same as that of the bullet. Kinetic energy, by the way, is energy of motion, things that move.
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