Ask question

Ask question

All categories

3 years ago

15

Before colliding, the momentum of block A is -100 kgm/, and block B is -150 kgm/s. After, block A has a momentum -200 kg*m/s.

What is the momentum of block B afterward? PLEASE HELP

1 answer:

rjkz [21]3 years ago

6 0

Answer:

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

Explanation:

Given

Before collision:

Momentum of block A = $p_{A1}$ = $-100\ kg\ ms^{-1}$

Momentum of block B = $p_{B1}$ = $-150\ kg\ ms^{-1}$

After collision:

Momentum of block A = $p_{A2}$ = $-200\ kg\ ms^{-1}$

Applying law of conservation of momentum to find momentum of block B after collision $p_{B2}$ .

$p_{A1}+p_{B1}=p_{A2}+p_{B2}$

Plugging in the given values and simplifying.

$-100-150=-200+p_{B2}$

$-250=-200+p_{B2}$

Adding 200 to both sides.

$200-250=-200+p_{B2}+200$

$-50=p_{B2}$

∴ $p_{B2}=-50\ kg\ ms^{-1}$

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

You might be interested in

What happens to the temperature of a substance while it is changing state?

MArishka [77]

It stays constant, because it's using that energy to change state

3 0

3 years ago

Read 2 more answers

A policeman is chasing a criminal across a rooftop at 10 m/s. He decides to jump to the next building which is 2 meters across f

valina [46]

Answer:

They will meet at a distance of 7.57 m

Given:

Initial velocity of policeman in the x- direction, $u_{x} = 10 m/s$

The distance between the buildings, $d_{x} = 2.0 m$

The building is lower by a height, h = 2.5 m

Solution:

Now,

When the policeman jumps from a height of 2.5 m, then his initial velocity, u was 0.

Thus

From the second eqn of motion, we can write:

$h = ut + \frac{1}{2}gt^{2}$

$h = \frac{1}{2}gt^{2}$

$2.5 = \frac{1}{2}\times 10\times t^{2}$

t = 0.707 s

Now,

When the policeman was chasing across:

$d_{x} = u_{x}t + \frac{1}{2}gt^{2}$

$d_{x} = 10\times 0.707 + \frac{1}{2}\times 10\times 0.5 = 9.57 m$

The distance they will meet at:

9.57 - 2.0 = 7.57 m

8 0

3 years ago

A subway train accelerates from rest at one station at a rate of 1.30 m/s^2 for half of the distance to the next station, then d

Minchanka [31]

This problems a perfect application for this acceleration formula:

   Distance = (1/2) (acceleration) (time)² .

During the speeding-up half:   1,600 meters = (1/2) (1.3 m/s²) T²
During the slowing-down half: 1,600 meters = (1/2) (1.3 m/s²) T²

Pick either half, and divide each side by 0.65 m/s²:

   T² = (1600 m) / (0.65 m/s²)

   T = square root of (1600 / 0.65) seconds

Time for the total trip between the stations is double that time.

   T = 2 √(1600/0.65) = <em>99.2 seconds</em> (rounded)

7 0

3 years ago

4.0 kg objects mive a distance of 7.8 m under the action of a cinstant gorce of 5.6 N. how much work is dine on object?

alexdok [17]

Answer:

43.68 J

Explanation:

Distance moved= 7.8 m

Force = 5.6 N

Work Done= Distance moved * Force

= 7.8 *5.6

=43.68 Joules

6 0

3 years ago

A cruise ship made a trip to Guam and back. The trip there took 12 hours and the trip

vekshin1

15 because i did the math, had this assignment, and also watch a video

6 0

3 years ago