100 kg 2m/s Two people collide head-on into cach other as they walk down the hal
ID: 2041276 • Letter: 1
Question
100 kg 2m/s Two people collide head-on into cach other as they walk down the hallway as illustrated in FIGURE 7-1A One has a mass of 50 kg and is moving at 2 m/s. The other has a mass of 100 kg and is also moving at 2 m/s. Which person applies the greater impulse in this interaction? The more massive person The less massive person c. Neitther- the impulse is the same Assume the impending collision illustrated in FIGURE 7-1A is perfectly inelastic, after collision how fast will the more massive person be moving? a. 0.67 m/s b. 0 m/s c 1 m/s d. 4 m/s A 25 kg anvil weighing 245 N is dropped to the floor from the top of my desk. Its speed right before impact is 5 m/s. How much force did the floor apply in stopping the anvil? a. 0 N b.) 245 N 490 N Not enough information given ?. d.Explanation / Answer
after collision speed of 100kg person is v1 = (m1-m2)*u1/(m1+m2) + (2*m2*u2)/(m1+m2)
v1 = ((100-50)*2/(100+50))-((2*50*2)/(100+50)) = -0.67 m/s
speed of 50 kg person is v2 = (2*m1*u1)/(m1+m2) + (m2-m1)*u2/(m1+m2)
v2 = ((2*100*2)/(100+50)) - ((50-100)*2/(100+50)) = 3.33 m/s
change in momentum of 100 kg person = (-100*0.67)-(100*2) = -267 kg m/s
change in momentum of 50 kg person = (50*3.33)-(50*2) = 66.5 kg m/s
Since impulse is directly proportional to the change in momentum
So
since 100 kg person's momentum changes more than 50 kg,there fore answer is
The more massive person
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since this is perfectly inelastic collision
then
m1*u1 - m2*u2 = (m1+m2)*V
(100*2)-(50*2) =(100+50)*V
V = 0.67 m/s is the correct answer
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if the floor applies the force which is equal to weight of anvil,then
the body will come to rest
so the answer is b) 245 N
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