some physics help.... please?????

Question

a few exam questions that are throwing me...
A 480 kg car moving at 14.4 m/s hits from behind another car moving at 13.3 m/s in the same direction. if thee second car has a mass of 570 kg and a new speed of 17.9 m/s what is the velocity of the first car after the collision?

A child pulls on a wagon handle at an angle 37° above the horizonatal with a force of 45N. if the wagon accelerates at 8.1m/s^2 horizontally, what is the mass of the wagon?

A person who normally weighs 200 lbs is standing on a scale inside an elevator. the elevator is moving upwards with a speed of 7 m/s, then begins to slow down at a rate of 5 m/s^2. before the elevator begins to slow down, the reading of the scale is ? and while the elevator is slowing down the reading of the scale is?
The choices are:
less than 200lbs, 100lbs
greater than 200lbs, 100lbs
200lbs, 100lbs
greater than 200lbs, 0lbs
none

THANKS

Answer 1

> A 480 kg car moving at 14.4 m/s...

This is a "conservation of momentum" problem. The total momentum before the collision must equal the total momentum after the collision.

Initial total momentum:
(m1)(v1_initial) + (m2)(v2_initial)

Final total momentum:
(m1)(v1_final) + (m2)(v2_final)

Write an equation setting those two equal:
(m1)(v1_initial) + (m2)(v2_initial) = (m1)(v1_final) + (m2)(v2_final)

They give you all the numbers except v1_final:
m1 = 480 kg
v1_initial = 14.4 m/s
v1_final = ?
m2 = 570 kg
v2_initial = 13.3 m/s
v2_final = 17.9 m/s

All you have to do is plug those numbers into the equation and then solve for v1_final.

> A child pulls on a wagon handle at an angle 37°

Use Fnet=ma; and consider horizontal and vertical components of force separately.

Since the acceleration is horizontal, the net force must also be horizontal. That means, all the vertical forces cancel out; so essentially we don't have to think about those.

The horizontal forces are:
1. The horizontal component of the force exerted by the child. This is: (45N)(cos37)
2. The force of friction pulling in the opposite direction.

We will have to assume #2 (friction) is zero, since there is not enough information in the problem to solve it otherwise.

So, the net force is horizontal, equal to (45N)(cos37):

Fnet = (45N)(cos37)
and:
Fnet = ma
so:
(45N)(cos37) = ma

They tell you how much "a" is. Solve for "m".

> A person who normally weighs 200 lbs ....

> before the elevator begins to slow down, the reading of the scale is ?

Before the elevator slows down, the person is not accelerating. That is: a=0. Therefore, the net force (Fnet) on the person must also be zero (since Fnet=ma).

Fnet = 0 (before slowdown)

Now, the total forces acting on the person are:
1. His weight (200 lbs. downward)
2. The upward pressure of the scale on his feet (call this F_scale). This also equals the reading on the scale.

Since Fnet = 0, these two forces must exactly cancel out; i.e. they're the same amount but in opposite directions:

Weight = F_scale (before slowdown)

That means F_scale (the reading on the scale) is 200 lbs. before the slowdown happens.

> Then begins to slow down at a rate of 5 m/s^2.

You can use Fnet = ma again. This time, "a" is not zero:

Fnet = (m)(5m/s²)

Again, the net force is a combination of the downward force of gravity (Weight); and the upward pressure of the scale (F_scale):

Fnet = Weight − F_scale

Combine the two equations:

(m)(5m/s²) = Weight − F_scale

Now use the fact that Weight = mg:

(m)(5m/s²) = mg − F_scale

Solve for F_scale:

F_scale = mg − (m)(5m/s²)
= m(g − 5m/s²)