Question

A 5.40  kg mess kit sliding on a frictionless surface explodes into two 2.70  kg parts, one moving at 3.50  m/s , due North, and the other at 5.50  m/s , 29.0  degrees North of East.

1.) What is the original speed of the mess kit?

2.) In what direction was the mess kit originally traveling? Give your answer as an angle in degrees measured CCW from due East (when looking Down); the size of the angle should be less than 180 degrees. If the angle would be larger than 180 degrees, instead measure CW from due East and give your answer as a negative angle.

3.) How much kinetic energy was added to the mess kit by the explosion?

Answer 1

using conservation of momentum :

(Mv )_before = m1v1 + m2v2

taking x-direction :

Mvx = m1v1x + m2v2x

5.40*vx = 0 + 2.70*(5.50*cos29)

vx = 2.41 m/s

taking y-direction :

M*vy = m1v1y + m2v2y

5.40*vy = 2.7(3.5) + 2.7(5.5sin29)

vy = 3.1 m/s

v = sqrt(vx^2 + vy^2)

v = 3.91 m/s

b)

theta = tan^-1(vy/vx)

theta = tan^-1(3.1/2.41)

theta = 52.14 degree north to east

c)

KE = KEfinal - KEintial

KE = 1/2*2.7*5.5^2 + 1/2*2.7*3.5^2 - 0.5*5.40*3.91^2= 10.85 J

KE = 57.375 - 41.28

KE = 16.1 J