A 3.00kg block of wood rests on the muzzle opening of a verticallyoriented rifle
ID: 1667342 • Letter: A
Question
A 3.00kg block of wood rests on the muzzle opening of a verticallyoriented rifle, the stock of the rifle being firmly planted on theground. When the rifle is fired, an 8.00g bullet (velocity=8.00*102 m/s, straight upward) becomes completelyembedded in the block. a) Using the conservation of linearmomentum find the velocity of the block/bullet system immediatelyafter the collision. b) Ignoring air resistance, determinehow high the block/bullet system rises above the muzzle opening ofthe rifle.answer: a)2.13m/s up, b)0.231m
Explanation / Answer
A 3.00kg block of wood rests on the muzzle opening of a verticallyoriented rifle, the stock of the rifle being firmly planted on theground. When the rifle is fired, an 8.00g bullet (velocity=8.00*102 m/s, straight upward) becomes completelyembedded in the block. a) Using the conservation of linearmomentum find the velocity of the block/bullet system immediatelyafter the collision. b) Ignoring air resistance, determinehow high the block/bullet system rises above the muzzle opening ofthe rifle. (a) Momentum conservation: m1v1+m2v2 = m1v1'+m2v2' Since the bullet is embedded in the block, they will move as oneafter collision, and we can rewrite the equation as m1v1+m2v2 =(m1+m2)V' Plug in values: 0.008(kg) x 8 x 102 + 3x0 = (3+0.008)V' V' = 2.13 m/s (b) Energy conservation: after the bullet is embedded into theblock, moving at 2.13m/s, which means it has kinetic energy(0.5mv2) . When the system rises to its highest point(the system stops), at this point, the system has potential energy(mgh) Therefore it is the conversion of Kinetic to Potential energy 0.5mv2 = mgh (both sides has 'm' we can remove this bydividing m on both side) 0.5 (2.13)2 = 9.8 x h h = 0.231m
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.