A marksman holds a rifle with a mass of 2.57 kg loosely, so it can recoil freely. He fires a bullet with a mass of 3.65 g horizontally with a velocity relative to the ground of 318.4 m/s. What is the final kinetic energy in J of the rifle

KE = 0.262 J

Explanation:

given data

rifle with  mass  Mr = 2.57 kg

bullet with mass Mb = 3.65 g

velocity relative to the ground Vb = 318.4 m/s

solution

we get here final KE of rifle that is express as

KE = 0.5 × Mr × (Vr)²    ..............1

here Vr is

Mb × Vb = Mr × Vr         .........2

put here value

3.65 ×  × 318.4 = 2.57 × Vr

Vr = 0.452 m/s

now put value in eq 1 we get

KE = 0.5 × 2.57 × (0.452)²

KE = 0.262 J

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s