Ch3-HW4-solutions - louey(cal2859 Ch3-HW4 florin(56930 This...

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louey (cal2859) – Ch3-HW4 – florin – (56930) 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points A bullet of mass m traveling horizontally at a very high speed v embeds itself in a block of mass M that is sitting at rest on a nearly frictionless surface. What is the final speed v f of the block-bullet system after the bullet embeds itself in the block? 1. v f = parenleftbigg m 2 m + M parenrightbigg v 2. v f = parenleftbigg M m parenrightbigg v 3. v f = parenleftBig m M parenrightBig v 4. v f = parenleftbigg M m + M parenrightbigg v 5. v f = parenleftbigg m m + M parenrightbigg v correct Explanation: Assume the bullet travels in the + x direc- tion, so that v bullet , i = ( v, 0 , 0 ) m / s . Define the system to be the bullet and the block. Assume that the net external force on the system is zero. Apply the momentum principle with the initial momentum being before the collision and the final momentum being after the collision. vectorp bullet , i + vectorp block , i = vectorp bullet , f + vectorp block , f mvectorv bullet , i + 0 = mvectorv f + Mvectorv f mvectorv bullet , i + 0 = ( m + M ) vectorv f vectorv f = parenleftbigg m m + M parenrightbigg vectorv bullet , i vextendsingle vextendsingle vectorv f vextendsingle vextendsingle = parenleftbigg m m + M parenrightbigg vextendsingle vextendsingle vectorv bullet , i vextendsingle vextendsingle = parenleftbigg m m + M parenrightbigg v. 002(part1of2)10.0points A car of mass 2800 kg collides with a truck of mass 4300 kg, and just after the collision the car and truck slide along, stuck together. The car’s velocity just before the collision was vectorv car , i = ( 40 , 0 , 0 ) m / s , and the truck’s velocity just before the colli- sion was vectorv truck , i = (− 13 , 0 , 31 ) m / s . The velocity of the stuck together car and truck just after the collision will be of the form vectorv sys , f = ( v sys , f ,x , 0 ,v sys , f ,z ) m / s . Find the x component, v sys , f ,x . Correct answer: 7 . 90141 m / s. Explanation: We can use the momentum principle. We know that the momenta before and after the collision will be given by vectorp sys , i = m car vectorv car , i + m truck vectorv truck , i vectorp sys , f = ( m car + m truck ) vectorv sys , f The momentum principle predicts that vectorp sys , i = vectorp sys , f , so we can use this to solve for vectorv sys , f . vectorp sys , f = ( m car + m truck ) vectorv sys , f vectorv sys , f = vectorp sys , i m car + m truck = m car vectorv car , i + m truck vectorv truck , i m car + m truck = ( 7 . 90141 , 0 , 18 . 7746 ) m / s . 003(part2of2)10.0points Find the z component, v sys , f ,z . Correct answer: 18 . 7746 m / s. Explanation: See the explanation for part 1.
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louey (cal2859) – Ch3-HW4 – florin – (56930) 2 004(part1of7)3.0points Object A has mass m A = 8 kg and initial momentum vectorp A,i = ( 19 , 4 , 0 ) kg · m / s , just before it strikes object B, which has mass m B = 11 kg. Just before the collision, object B has initial momentum vectorp B,i = ( 6 , 8 , 0 ) kg · m / s .
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