1. An average force of 1200 N acts on an object for 3.0 ms. The impulse imparted

Question

1. An average force of 1200 N acts on an object for 3.0 ms. The impulse imparted by this force to the object is:

a. 3600 N s

b. 3.6 N s

c. 400 N s

d. 2.5 N s

2. A baseball bat is in contact with the ball for 5.0 ms. During that contact time the impulse is 12.0 N s. The force applied by the bat to the ball is:

a. 2500 N

b. 60 N

c. 2.4 N

d. 0.06 N

3. A 1000 kg car moves initially at a constant speed of 20 m / s on a straight road. Its linear momentum is:

a. 50 kg m / s

b. 100 kg m / s

c. 20,000 kg m / s

d. 20 kg m / s

4. A ball of 0.150 kg initially moves towards a wall at 10 m / s. Then it bounces moving in the opposite direction at a speed of 8 m / s. The magnitude of your change in momentum is:

a. 3.7 kg m / s

b. 1.5 kg m / s

c. 1.2 kg m / s

d. 2.7 kg m / s

5. The graph of force versus time shows us how a force acts on a particle during a time interval:

The impulse that this force imparts to the particle is:

a. 3.5 N s

b. 37.5 N s

c. 40 N s

d. 2.5 N s

Explanation / Answer

here,

1)

force , F =1200 N

time taken , t = 3 ms = 0.003 s

the impulse , I = F * t

I = 1200 * 0.003 kg.m/s = 3.6 kg.m/s

the correct option is b) 3.6 N.s

2)

time interval , t = 5 ms = 0.005 s

impulse , I = 12 N.s

the force applied , F = I /t

F = 12 /0.005 N

F = 2500 N

the correct option is a) 2500 N

3)

mass , m = 1000 kg

speed ,v = 20 m/s

the linear momentum , P = m * v

P = 20000 kg.m/s

the correct option is c) 20000 kg.m/s

4)

mass , m = 0.15 kg

initial speed , u = - 10 m/s

final speed , v = 8 m/s

the change in momentum , dP = m * ( v - u)

dP = 0.15 * ( 8 - ( -10)) kg.m/s

dP = 2.7 kg.m/s

the correct option is d_) 2.7 kg.m/s

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