Question Video: Using the Conservation of Momentum to Find Unknown Velocity

A rocket of mass 200.0 kg in deep space moves with a velocity of (121𝑖 + 38.0𝑗) m/s. Suddenly, it explodes into three pieces. The first piece has a mass of 78.0 kg and moves at (βˆ’321𝑖 + 228𝑗) m/s. The second piece has a mass of 56.0 kg and moves at (16.0𝑖 βˆ’ 88.0𝑗) m/s. Find the velocity of the third piece.


Video Transcript

A rocket of mass 200.0 kilograms in deep space moves with a velocity of 121𝑖 plus 38.0𝑗 metres per second. Suddenly, it explodes into three pieces. The first piece has a mass of 78.0 kilograms and moves at negative 321𝑖 plus 228𝑗 metres per second. The second piece has a mass of 56.0 kilograms and moves at 16.0𝑖 minus 88.0𝑗 metres per second. Find the velocity of the third piece.

We can label this velocity 𝑣 sub three. And we know it will have two components, an 𝑖 or π‘₯- and 𝑗 or 𝑦-component. Considering the information we’ve been given, we’ll label the mass of the rocket as capital 𝑀. And we’ll call its velocity 𝑣 sub 𝑀. We’ll label the first piece of exploded mass 𝑀 sub one and the velocity of that mass 𝑣 sub one. And we’ll call the second piece that exploded 𝑀 sub two. And the velocity of that piece we’ll label 𝑣 sub two.

If we drew a sketch of the initial and final states of this scenario, initially, we have our large rocket mass capital 𝑀 moving at a velocity 𝑣 sub 𝑀. And finally, after the explosion, our big mass 𝑀 is broken up into three smaller pieces each moving with their own velocity. To solve for the velocity of the third unknown piece, we’ll use the principle of the conservation of momentum. This principle tells us that if we add up all the momentum in our system initially, that’s equal to all the momentum in our system finally. And because momentum is a vector, that means that this conservation applies independently to every dimension of our situation.

In our case, we have a two-dimensional scenario, whose dimensions we can call π‘₯ and 𝑦. We can consider these dimensions separately one at a time. We’ll start in the π‘₯-direction and write out the conservation of linear momentum in that dimension. We can say that the initial momentum in the π‘₯-direction is equal to the final momentum of our system in that same direction. Knowing that the momentum of an object is equal to its mass times its velocity, we can write that the initial momentum of our system in the π‘₯-direction is the mass capital 𝑀 times the speed of that mass in the π‘₯-direction.

When we consider the momentum in the π‘₯-direction finally after a rocket has broken into three separate pieces, that’s equal to π‘š one times the π‘₯-component of π‘š one’s velocity plus π‘š two times 𝑣 two π‘₯ plus π‘š three times 𝑣 three π‘₯, that is, the sum of the momenta of each of our three separate masses in the π‘₯-direction. In the problem statement, we’re told the value for capital 𝑀, π‘š sub one, and π‘š sub two. We can use these three values to solve for π‘š sub three. π‘š sub three is equal to capital 𝑀 minus π‘š sub one minus π‘š sub two or plugging in for the given values 200.0 kilograms minus 78.0 kilograms minus 56.0 kilograms. This equals 66.0 kilograms which we’ll keep track of as the mass of our third particle.

We now have all the information needed in order to solve for the π‘₯-component of our third piece 𝑣 sub three π‘₯. We rearrange this equation to solve for 𝑣 sub three π‘₯. When we plug in for these many values, we’re inserting the values of the masses of our original and final three pieces as well as the π‘₯-component of the velocities of our original piece and two of our final pieces. In terms of the units involved, notice that the unit of kilograms cancel out and we’ll be left with a result in metres per second. When we calculate this fraction to three significant figures, we find a result of 732 metres per second. So we now know the π‘₯-component of the velocity of our third piece, 732𝑖 metres per second.

We next wanna solve for the 𝑗 component of that velocity. To do this, we’ll follow a similar process. In the 𝑦-direction, we know that the initial momentum is equal to the final momentum. Initially, the momentum in this direction is capital 𝑀 times 𝑣 sub 𝑀 in the 𝑦-dimension. And finally, the 𝑦 momentum is π‘š one times 𝑣 one 𝑦 plus π‘š two times 𝑣 two 𝑦 plus π‘š three times 𝑣 three 𝑦. Once again, we rearrange this expression to solve for the variable we’re interested in. In this case, the 𝑦-component of particle three is velocity. When we plug all these values in this time using the 𝑦-component of our velocities is for our original mass and two of our final masses, we find that to three significant figures 𝑣 sub three 𝑦 is negative 79.6 metres per second. We now have an expression for the complete two-dimensional velocity of the third piece of exploded mass.

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