An object with total mass m total = 16.2 kg is sitting at rest when it explodes

Question

An object with total mass mtotal = 16.2 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.6 kg moves up and to the left at an angle of 1 = 23° above the –x axis with a speed of v1 = 25.2 m/s. A second piece with mass m2 = 5.1 kg moves down and to the right an angle of 2 = 28° to the right of the -y axis at a speed of v2 = 23.6 m/s.

1)What is the magnitude of the final momentum of the system (all three pieces)?

__________________kg-m/s

2)What is the mass of the third piece?

________________kg

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

______________m/s

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

_________________m/s

5)What is the magnitude of the velocity of the center of mass of the pieces after the collision?

_________________m/s

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

_______________J

Explanation / Answer

1. explosion happens due to internal forces.

hence momentum will be same.

so final momentum = initial momentum = 0

2. M = m1 + m2 + m3

16.2 = 4.6 + 5.1 + m3

m3 = 6.5 kg

3.
v1 = 25.2 (- cos23i + sin23j) = - 23.2i + 9.85 j m/s

v2 = 23.6 ( sin28i - cos28j) = 11.08i - 20.8 j m/s

applying momentum conservation,

0 = m1 v1 + m2v2 + m3v3

0 = 4.6 ( - 23.2i + 9.85 j ) + 5.1 ( 11.08i - 20.8 j ) + 6.5v3

v3 = 7.72i + 9.35j m/s

v3x = 7.72 m/s

4.
v3y = 9.35 m/s

5. velocity of centre of mass will not change, it will remain zero.

6. KE = m1 v2^2 /2 + m2 v2^2 /2 + m3 v3^2 / 2

= (4.6 x 25.2^2 / 2) + ( 5.1 x 23.6^2 / 2) + (6.5 x (7.72^2 + 9.35^2) / 2)

= 3358.66 J

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