a. m1*V1 = m2*V2,
0.146*V1 = 0.00337*15238 = 51.35,
V1 = 351.7 m/s.
b. KE = 0.5m*V^2
KE = 0.5*0.00337*(15238)^2 = 391,251 J.
c. KE = 0.5*0.146*(351.7)^2 = 9.030 J.
A pitcher claims he can throw a 0.146 kg
baseball with as much momentum as a 3.37 g
bullet moving with a speed of 15238 m/s.
What must be its speed if the pitcher’s
claim is valid?
Answer in units of m/s
What is the kinetic energy of the bullet?
Answer in units of J
What is the kinetic energy of the ball?
Answer in units of J
1 answer