f5ch10 - 266 CHAPTER 10 COLLISIONS CHAPTER 10 Answer to...

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266 CHAPTER 10 COLLISIONS C HAPTER 10 Answer to Checkpoint Questions . ( a ) unchanged; ( b ) unchanged; ( c ) decreased . ( a ) zero; ( b ) positive; ( c ) positive direction of y . ( a ) kg m/s; ( b ) kg m/s; ( c ) J . ( a ) ; ( b ) kg m/s . ( a ) kg m/s; ( b ) kg m/s; ( c ) kg m/s . ( a ) kg m/s; ( b ) kg m/s . ( a ) increases; ( b ) increases Answer to Questions . all tie . ( a ) same (zero); ( b ) decreases; ( c ) decreases; ( d ) same (zero); ( e ) decreases; ( f ) decreases . b and c . ( a ) increases (with a limit of initial height h = : cm); ( b ) zero; ( c ) h = h = cm . ( a ) one stationary; ( b ) ; ( c ) ; ( d ) equal (pool player's result) . ( a ) ; ( b ) ; ( c ) ; ( d ) ; ( e ) less . ( a ) and tie, then and tie; ( b ) ; and tie; then . a: m/s, leftward; f and g: m/s, rightward; other blocks: . ( a ) rightward; ( b ) rightward; ( c ) smaller . ( a ) positive; ( b ) positive; ( c ) and . positive direction of x axis
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CHAPTER 10 COLLISIONS 267 . ( a ) lower (almost zero); ( b ) greater (baseball may smash into the ceiling) E ( a ) The magnitude of the force is F = p t = : kg m/s s = N : ( b ) The speed of the car increases by v = p m = : kg m/s kg = : m/s : E If F is the magnitude of the average force then the magnitude of the impulse is J = F t , where t is the time interval over which the force is exerted (see Eq. { ). This equals the magnitude of the change in the momentum of the ball and since the ball is initially at rest it equals the magnitude of the nal momentum mv . When F t = mv is solved for v the result is v = F t=m = ( N)( s) = ( : kg) = : m/s. E The magnitude of the force is F = j p j m = m j v j t = ( kg)( m/s) : s = : N : E ( a ) The magnitude of the change in momentum for the ball is j p j = mv ( mv ) = mv . Thus the magnitude of the average force exerted on the wall is F = j p j t = mv t : ( b ) Numerically, F = ( kg)( : m/s) : s = N :
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268 CHAPTER 10 COLLISIONS E The magnitude of the average force is F = j p j t = m j v j t = ( : kg)( m/s m/s) : s = : N ; which is equivalent to lb. E As he reaches the surfrace of the water, La Mothe (with mass m ) has reached a speed v , which sati es mgh = mv , where h = ft : As he decelerates in the water under the in uence of an average force F , his average speed is v = v= . The time it takes for him to stop from the speed v is therefore t = l= v = l=v , where l = in. is the depth of the water. From F t = p = mv we then obtain F = mv t = mv l=v = m ( gh ) l = ( lb)( ft) ( in.)( : ft = in.) = : lb : E The maximum deceleration a m that the paratrooper (of mass m and initial speed v ) could take is given by a m = F m =m , where F m = : N. Using v = a m d m , we nd the minimum depth of snow: d m = v a m = mv F m = ( kg)( m/s) ( : N) = : m : E Take the initial direction of motion to be positive and let F be the magnitude of the average force, t be the duration of the force, m be the mass of the ball, v i be the initial velocity of the ball, and v f be the nal velocity of the ball. Then the force is in the negative direction
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This note was uploaded on 01/22/2012 for the course PHYS 2101 taught by Professor Grouptest during the Fall '07 term at LSU.

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f5ch10 - 266 CHAPTER 10 COLLISIONS CHAPTER 10 Answer to...

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