Laws of conservation_10_CHD_ worksheet

Worksheet 1( momentum and impulse)

Name………………………………………………class……………………………….Date…………

1.      Maloney threw a fastball. Before the ball reached the plate, Alex took off his glove; he then caught the pitch barehanded.

a)      Determine the impulse required to stop a 0.145-kg baseball moving at 35.7 m/s.

b)       If this impulse is delivered to the ball in 0.020 seconds, then what is magnitude of the force acting between the bare hand and the ball?

2.      Kara Less was applying her makeup when she drove into South's busy parking lot last Friday morning. Unaware that Lisa Ford was stopped in her lane 30 feet ahead, Kara rear-ended Lisa's rented Taurus. Kara's 1300-kg car was moving at 11 m/s and stopped in 0.14 seconds.

a.       Determine the momentum change of Kara's car.

b. Determine the impulse experienced by Kara's car.

c. Determine the magnitude of the force experienced by Kara'

3.      An impulsive force is applied on a body of mass 3 kg.

a.       If the time of force application is 5s find the impulse of the box given below.

b.       Find applied force which makes 10m/s change in the velocity of the box in 5s if the mass of the box is 4kg. 

4.      An object travels with a velocity 4m/s to the east. Then, its direction of motion and magnitude of velocity are changed. Picture given below shows the directions and magnitudes of velocities. Find the impulse given to this object. 

5.      Find the impulse and force which make 12m/s change in the velocity of object having 16kg mass in 4 s.

6.       Applied force vs. time graph of object is given below. Find the impulse of the object between 0-10s.

7.      A ball having mass 500g hits wall with a10m/s velocity. Wall applies 4000 N force to the ball and it turns back with 8m/s velocity. Find the time of ball-wall contact.

Answers

1.      a. -5.18 N•s
b. 260 N (rounded from 259 N)

2.      a) -1.4 x 10⁴ kg•m/s
b) -1.4 x 10⁴ N•s
c) 1.0 x 10⁵ N

3.       

a.       Impulse=Force.Time Interval

Impulse=15N.5s

Impulse=75N.s

b.       

Impulse=Change in momentum

F.t=p₂-p₁

F.t=m. (V₂-V₁)

F.t=4kg.10m/s=40kg.m/s Impulse of the box is 40kgm/s

F=40kg.m/s/5s=8N Applied force

4.       

I=F.Δt=Δp=m.ΔV

where ΔV=V₂-V₁=-3-4=-7m/s

I=m.ΔV=3.(-7)=-21kg.m/s

5.       

F.Δt=ΔP=m.ΔV

F.4s=16kg.12m/s

F=48N

F.Δt=Impulse=192kg.m/s

6.      Area under the force vs. time graph gives us impulse.

F.Δt=20.2/2+20.(6-2)+20.(10-6)/2

F.Δt=140kg.m/s

7.      F.Δt=ΔP=m.ΔV=m.(V₂-V₁)

-4000.Δt=0,5kg.(8-10)

Δt=0,00025s