Momentum & Impulse. To increase an objects momentum, either increase its or its. Momentum is a, so DIRECTION is a necessity.

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1 Momentum: p mv Momentum & Impulse To increase an objects momentum, either increase its or its. Momentum is a, so DIRECTION is a necessity. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Newton s original version of what is now known as his 2 nd Law of Motion was not F = ma, but rather that the rate of change of momentum of an object is proportional to the net force applied AND the momentum and the net force are in the same direction. Write out (in equation form) Newton s original intent for the 2 nd Law ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Impulse: F t vector N-s mass velocity p/t = F The impulse of a force has units of, and graphically it is the area under a F-t graph. Impulse-Momentum Theorem: F t p or F t p m mv1 or FAt mav (Ft=mv) Longer stopping time less force required to stop an object Shorter stopping time more force required to stop an object The amount that momentum changes depends on two factors: Contact time Contact Force and 1) Egg dropping on pillow vs. concrete. Examples to think about 2) Follow thru on a golf swing. 3) Recoil on a firing gun. 4) Hitting a baseball with a bat vs. hitting a tennis ball with a racket.

2 Momentum & Impulse Problems SOLUTIONS 1. A 50 kg object is moving east at 10 m/s. What is its momentum? p mv (50kg)(10 m s [E]) 500N s [E] OR 500N s [E] note : N s kgm s (both sets of units are thesame!!!!!) 2. A 30 kg object moving west at 5 m/s turns around and heads east at 10 m/s. Find its change in momentum while turning? v 1 5 m s [W ] 5 m s 10 m s [E] 10 m s p p 2 p 1 p m mv 1 m( v 1 ) p (30kg)[10 m s (5 m s )] (30kg)(15 m s ) 450N s[e] or 450kg m s [E] p1 mv1 150N s [ E] + p = p2 mv2 300N s [ E] p = 450 N-s 3. A 200 kg child is sitting at rest in a wagon. What velocity will the child acquire if the wagon is pushed by a westward 100N force for 3 seconds? Ft p Ft m mv 1 Ft mv 1 m (100N[W ])(3sec) (200kg)(0) 200kg 1.5 m s [W ]

3 4. What force acting for sec. will change the velocity of a 0.1 kg baseball from 30 m/sec toward home plate to 40 m/sec toward centerfield? v 1 30 m s [toward home] 30 m s 40 m s [towards outfield] 40 m s Ft p Ft m mv 1 F m mv 1 t F (0.1kg)(40 m s ) (0.1kg)(30 m s ) 0.001sec 7,000N[towards CF] Notice that the sign of the answer is dependant upon the direction of the coordinate system that you used. I chose "toward home" as neagtive. If I chose "toward the outfield" as negative, the sign on my force would have been negative. This is why, to alleviate any sign issues, I gave my answer as "towards the outfield" instead of + or What force acting for 1.5 sec. is needed to give a 7.27 kg bowling ball a velocity of 6 m/sec? Ft p Ft m mv 1 F m mv 1 t F (7.27kg)(6 m s ) (7.27kg)(0 m s ) 1.5sec 29.1N [forward] 6. A bullet having a mass of 0.05 kg moving with a velocity of 400 m/sec penetrates a distance of 0.1 meters into a wooden block firmly attached to the Earth. Calculate a. the acceleration of the bullet b. the accelerating force c. time of impact d. the impulse of the collision Since a displacement is given, we can usean equation of motion tosolvefor t!!!! x 1 2 t( v 1 ) t(0 400) t sec.0005sec 0.5 milli sec v 1 at a(.0005) a m s 2 800,000 m s 2 F ma (0.05kg)( m s 2 ) N 40,000N impulse Ft ( N)( sec) 20N s

4 7. A wagon having a mass of 1000 kg is pushed by a constant force for 56 seconds. Its speed increased from 4 m/sec to 32 m/sec. What is the wagon s change in momentum? What force pushed the wagon? p m mv 1 m( v 1 ) (1,000kg)(32 m 4 m s s ) 28,000N s [forward] Ft p F p t 28,000N s 500N [forward] 56sec 8. Calculate the momentum of a golf ball of mass 60 grams moving with a velocity of 70 m/sec. p mv (.060kg)(70 m s ) 4.2kg m s OR 4.2N s 9. A measurement of the momentum of a proton yields a value of 6.8 x kg m/sec. The mass of a proton is 1.7 x kg. What is its velocity? p mv v p m 10. Find the impulse for each graph below. 6.8E 21N s 1.7E 27kg m s 4,000,000 m s a) b) F (N) F (N) t (s) t (s) impulse area a) impulse 1 (6sec)(40N) (10sec)(40N) 120N s 200N s 320N s b) impulse 1 (2sec)(30N) (6sec)(30N) (4sec)(20N) (4sec)(30N) (2sec)(50N) 30N s 180N s 160N s 100N s 470N s 11. Explain why, using the concept of impulse and momentum, an airbag in a car helps to reduce injuries. F t p Whether you are driving in a car with an airbag or without, you still go from v1 to v 2. Therefore, the change in momentum is the same for either case. Since an airbag increases the stoppingtime ( t), then theforce must go down tokeep the change in momentum the same. Therefore, airbags reduce the stoppingforce, thus helping to reduce injury. In other words. SMALL TIME x LARGE FORCE = CHANGE IN MOMENTUM OR LARGE TIME x SMALL FORCE = CHANGE IN MOMENTUM

5 12. Explain why, using the concept of impulse and momentum, the forces generated when a baseball player hits a baseball are so enormous, whereas the forces generated when a tennis player hits a tennis ball are much less. When a baseball player hits a baseball, the bat is very "hard", and sois Asa result, the time during which the bat is in contact with theball is very short. This short time would make the change in momentum very small unless the force was very high. very fast, the force must be VERY high to cause ON THE OTHER HAND Since baseballs are thrown very fast and the ball. (since F t p) then hit this large momentum change. When a ball hits a tennis racket, the racket flexes, and the contact time between the racket face and the ball is much more than with a ball and a bat. In order to give the same change in momentum to both balls, a smaller force is imparted to the tennis ball (compared to the baseball) because of this larger contact time.