# Exercise 5B| 1 | 2 | 3 | 4 | 5 | back to g solutions | physics home|

By: Ryan Carpenter

formulas

# EK = .5mv^2

1)       Calculate the speed of an 8.0E 4 kg airliner with a kinetic energy of 1.1E9 J

EK = .5mv^2

1.1E9 = .5 (8.0E4)v^2

1.1E9/(.5 (8.0E4)) = v^2

(1.1E9/(.5 (8.0E4)))^.5 = v

v = 165.8312395

v = 170 m/s      TOP

2)       What is the speed of a 0.145 kg baseball if its kinetic energy is 109 J?

EK = .5mv^2

109 = .5 (0.145)v^2

109/(.5 (0.145)) = v^2

(109/(.5 (0.145)))^.5 = v

v = 38.77432496

v = 38.8 m/s     TOP

3)       Two bullets have masses of 3.0 g and 6.0 g respectively.  Both are fired with a speed of 40.0 m/s.  (a) Which bullet has more kinetic energy? (b) what is the ratio of their kinetic energies?

Bullet 1

EK = .5mv^2

EK = .5(.03)(40)^2

EK = 24 j

Bullet 2

EK = .5mv^2

EK = .5(.06)(40)^2

EK = 48 j

Thus bullet 2 has a greater kinetic energy with a ration of  48:24 or 2:1.              TOP

4)      Two 3.0 g bullets are fired with speeds of 40.0 m/s and 80.0 m/s, respectively.  (a) What is their kinetic energies? (b) Which bullet has more kinetic energy? (c) What is the ration of their kinetic energies?

Bullet 1

EK = .5mv^2

EK = .5(.03)(40)^2

EK = 24 j

Bullet 2

EK = .5mv^2

EK = .5(.03)(80)^2

EK = 96 j

Bullet 2 has a greater kinetic energy, with ratio of 96:24 or 4:1   TOP

5)      A car has a kinetic energy of 4.32E5 J when traveling at the speed of 23 m/s.  What is its mass?

EK = .5mv^2

4.32E5 =  .5m(23)^2

4.32E5/.5(23)^2 = m

m = 1633.270321

m = 1600 kg    TOP