Question

An object with total mass mtotal = 14.9 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.8 kg moves up and to the left at an angle of θ1 = 23° above the –x axis with a speed of v1 = 27.9 m/s. A second piece with mass m2 = 5.1 kg moves down and to the right an angle of θ2 = 28° to the right of the -y axis at a speed of v2 = 20.9 m/s.

m3=5KG

What is the x-component of the velocity of the third piece?

What is the y-component of the velocity of the third piece?

Calculate the increase in kinetic energy of the pieces during the explosion.

Answer 1

Since the object was at rest therefore the momentum after the explosion should be zero by the conservation of momentum.
therfore
The third particle should balance the momentum of the both 1 and 2 particle .
We resolve the momentum of the first and second particle along the x axis
Therfore we get X component velocity of mass 3
M₃V₃Cos(theta) = M₁*V₁Cos23 + M₂V₂Sin28
M₃V₃Cos(theta) = 4.8*27.9Cos23 + 5.1*20.9*Sin28 = 173.314
Therfore
V₃Cos(theta) = 173.314/M₃ = 173.314/5 = 34.66 m/s
Similarly for Y axis component we get
M₃V₃Sin(teta) = 4.8*27.9Sin23 - 5.1*20.9*cos28 = -41.787
V₃Sin(theta) = - 41.787/5 = - 8.357 m/s
V₃ = ((V₃Sintheta)² +( V₃Cos(theta))²)^1/2
V₃ = 35.65 m/s
minus sign shows that our assumed direction is wrong.
therefore the m₃ is in the fourth quadrant instead of 1 what we assumed.
But since we want magnitude so it will not affect our answer.
Increase in kinetic energy
1 particle = (1/2)mV₁² = 0.5*4.8*27.9² = 1868.184 J
2 Particle = 0.5*5.1*20.9² = 1113.86 J
3 Particel = 0.5*5*(35.65²) = 3177.88 J