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An object with mass $m,$ initially at rest, explodes into two fragments, one with mass $m_{A}$ and the other with mass $m_{B},$ where $m_{A}+m_{B}=m .$ (a) If energy $Q$ is released in the explosion, how much kinetic energy does each fragment have immediately after the collision? (b) What percentage of the total energy released does each fragment get when one fragment has four times the mass of the other?

(a) $K_{A}=\left(\frac{m_{B}}{m_{A}+m_{B}}\right) Q \quad-\quad K_{B}=\left(\frac{m_{A}}{m_{A}+m_{B}}\right) Q$(b) $\frac{K_{A}}{Q}=20.0 \% \quad-\quad \frac{K_{B}}{Q}=80.0 \%$

Physics 101 Mechanics

Chapter 8

Momentum, Impulse, and Collisions

Moment, Impulse, and Collisions

Cornell University

University of Michigan - Ann Arbor

University of Winnipeg

McMaster University

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There is no external horizontal force on the object. Some momentum will be conserved so we can write B plus B B equals to zero. From here, we can write equals two minus pd. So now putting values we get area m maybe equals two minus of minus. M v v v. Okay, so from here we can write we b equals two. I am a deal whereby we I am a deal whereby MB multiplied by leak. Okay, See, this is exclusion number one No. The total energy released from the explosion explosion is the kinetic energy. So we can write q equals two total kinetic energy That is kinetic energy of a plus Kinetic energy of the Okay, so now we can write two equals two K plus one by two MB We be square No putting the value of will be In this equation we get two equals two a a plus one by two m b and we b s m a divided by m b multiplied by three a and holy square. So now after solving it, we get mhm two equals two and putting value of K A. That is half m A V a square half m a we a square plus one by two And solving this we gas m a v a square m a by M. V. This is the relations from here. By taking common, we get one by two m a v a holy sphere, and we get one plus m a d m v. Okay, so we can write. This is okay, so we can write q equals two e I am a plus MB They were by envy. Okay, so from here, the kinetic energy of a comes out to be m b divided by m A plus MB multiplied by Q So this is the answer for the kinetic energy of particles. A. Okay, so now mhm moving forward Substituting this equation we get q equals two m b d baby m a plus MB manipulated by q plus kv. Okay, so now from here we can write KB equals to one minus m b divided by m plus M v not a play by Q. So from here, after solving eight k b equals two m e divided by m A plus MB manipulated by. So this is the the kinetic energy of B in the equation for no for the part B. We have some values that is, we have m equals to four times off MB. So now putting all these situations, we get a equals two and be they were by is four times of M V c four MV plus MB. Thank you. So from here, we get Cube, I fight. So from here now we can right? Que que equals to 20.0%. This is the answer for the this part. Okay, So similarly, we can get K b equals two 100% minus K. Bye, Q, That is 80%. Okay, so KB by. So from here, we get KB by Q as 80%. Okay, so this is the answer for the part B, okay?

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