A large, lit m80 (a type of firework) is dropped off of a bridge. When the m80 is 12.30 m below the its drop point, it is moving with a speed of 12.30 m/s down. At this point the m80 explodes into two pieces, one piece with a mass of 0.265 kg moves downward with a speed of 18.90 m/s, and the other piece has a mass of 0.350 kg. If mass is conserved, what are the speed and direction of the 0.350 kg piece after the explosion? Ignore wind resistance.

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A large, lit m80 (a type of firework) is dropped off of a bridge. When the m80 is 12.30 m below the its drop point, it is moving with a speed of 12.30 m/s down. At this point the m80 explodes into two pieces, one piece with a mass of 0.265 kg moves downward with a speed of 18.90 m/s, and the other piece has a mass of 0.350 kg. If mass is conserved, what are the speed and direction of the 0.350 kg piece after the explosion? Ignore wind resistance.

Expert Solution
Step 1

Consider the following diagram,

Physics homework question answer, step 1, image 1

Step 2

Given the mass is conserved. Therefore the mass of lit M-80 will be the sum of masses of the two pieces. Let v be defined as the velocity of a piece having mass 0.350 kg, and the direction of the velocity be defined as θ. Let u be defined as the initial velocity, and it is defined as 12.30 m.s-1.

Using the conservation of momentum,

Mu=m1v1+m2v20.265 kg+0.350 kg12.30 m.s-1i^=0.265 kg18.90 m.s-1i^+0.350 kgvcosθi^+vsinθj^ m.s-17.5645i^=5.0085+0.350vcosθi^+0.350vcosθj^

Step 3

Comparing the x-component and y-component,

7.5645=5.0085+0.350vcosθ0.350vcosθ=2.556vcosθ=2.5560.350vcosθ=7.3028                .........1vsinθ=0                        ..........2

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