Physics, 11.10.2019 13:10 josecano2

1.

which of the following equations can be used to directly calculate an object s momentum, p?
(1 point)

p = mv

physa_u3_q2_question1
p = fdeltat
deltap = fdeltat

2.

what are the si units for momentum?
(1 point)

n·m
j
kg·m/s
kg·m/s2

3.
the change in an object s momentum is equal to (1 point)

the product of the mass of the object and the time interval.
the product of the force applied to the object and the time interval.
the time interval divided by the net external force.
the net external force divided by the time interval.

4.
the impulse experienced by a body is equivalent to the body s change in (1 point)

velocity.
kinetic energy.
momentum
force.

5.
which of the following statements properly relates the variables in the equation fdeltat = deltap? (1 point)

a large constant force changes an object s momentum over a long time interval.
a large constant force acting over a long time interval causes a large change in momentum.
a large constant force changes an object s momentum at various time intervals.
a large constant force does not necessarily cause a change in an object s momentum.

6.
a 75 kg person walking around a corner bumped into an 80 kg person who was running around the same corner. the momentum of the 80 kg person (1 point)

increased.
decreased.
remained the same.
was conserved.

7.
a 20 kg shopping cart moving at a velocity of 0.5 m/s collides with a store wall and stops. the momentum of the shopping cart (1 point)

increases.
decreases.
remains the same.
is conserved.

8.
a billiard ball collides with a second identical ball in an elastic head-on collision. what is the kinetic energy of the system after the collision compared with the kinetic energy before the collision? (1 point)

unchanged
one-fourth as great
two times as great
four times as great

9.
which of the following best describes the momentum of two bodies after a two-body collision if the kinetic energy of the system is conserved? (1 point)

must be less
must also be conserved
might also be conserved
is doubled in value

10.
a 90 kg halfback runs north and is tackled by a 120 kg opponent running south at 4 m/s. the collision is perfectly inelastic. just after the tackle, both players move at a velocity of 2 m/s north. calculate the velocity of the 90 kg player just before the tackle. (1 point)

10 m/s south
5 m/s north
10 m/s north
7.5 m/s south

11.
if two carts collide, what type of collision will conserve both momentum and kinetic energy? ” (1 point)

head on
perfectly inelastic
none
perfectly elastic

Answers: 2

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Answers

Answer from: hmatt5643

Answers

1)P = mv

2)Kg.m/s

3)The product of the force applied to the object and the time interval.

4)Momentum

5)A large constant force acting over a long time interval causes a large change in momentum.

6)Was conserved

7)Is conserved

8)Unchanged.

9)Must also be the same

10)10 m/s, north

11)Perfectly elastic.

Explanation

Momentum is the product of velocity and the mass. That is;

Momentum = velocity × mass . Where p -- momentum, P= mv

Since; momentum = velocity × mass, we can derive the S.I units.

Velocity – m/s

Mass – Kg.

So the SI units will be (Kg×m/s) = Kg.m/s

From the newton’s second law of motion,

F=Ma = (mv – mu)/t

Ft = mv – mu. Change in momentum is also equal to the product of force and time.

Impulse is the force acting on an object for a short time. It is calculated as;

Impulse = ft = change in momentum.

So the answer is momentum

When to bodies collide, the collision may result into an elastic or inelastic collision. In both cases the momentum will be conserved.

When the shopping cart collides with the wall, the momentum before colliding and after colliding will be the same. The momentum is always conserved.

There are two types of collision. Elastic and in-elastic collision. In an elastic collision both momentum and K.E are conserved while in an in-elastic collision only momentum is conserved. So the kinetic energy is unchanged.

At all cases of collision the momentum is always conserved. So is kinetic energy was conserved then, momentum must also have been conserved.

10)

Momentum before collision = momentum after collision

Taking the velocity towards north to be positive,

(90 × V) + (120×-4) = (90+120)× 2

90V – 480 = 210 × 2

90V = 420+480

V = 900/90

= 10 m/s

Since the answer is positive, the answer is 10 m/s, north.

11)

Answer from: Quest

The objects do not reach terminal velocity there is not enough time from the time the object falls from when it hits the ground to achieve terminal velocity. an object must be placed higher to achieve terminal velocity in order to be in free fall.

Answer from: Quest

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000

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