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Genrish500 [490]
3 years ago
5

A plastic ball of mass 0.200 kg moves with a velocity of 0.30 m/s. This plastic ball collides with a second plastic ball of mass

0.100 kg that is moving along the same line at a velocity of 0.10 m/s. After the collision, the velocity of the 0.100 kg ball is 0.36 m/s. What is the velocity of the second ball?
Physics
1 answer:
GrogVix [38]3 years ago
8 0
<h2>Velocity of 0.2 kg ball is 0.17 m/s</h2>

Explanation:

Let the mass of balls be m and M.

Initial velocity ball 1 be u₁, ball 2 be u₂

Final velocity ball 1 be v₁, ball 2 be v₂

Initial momentum = m x u₁ + M x u₂ = 0.2 x 0.3 + 0.1 x 0.1 = 0.07 kgm/s

Final momentum = m x v₁ + M x v₂ = 0.2 x v₁ + 0.1 x 0.36 = 0.2v₁ + 0.036

We have momentum conservation

Initial momentum = Final momentum

0.07 = 0.2v₁ + 0.036

v₁ = 0.17 m/s

Velocity of 0.2 kg ball is 0.17 m/s

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2 years ago
Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu
vlada-n [284]

Answer:

\frac{1}{10}M

Explanation:

To apply the concept of <u>angular momentum conservation</u>, there should be no external torque before and after

As the <u>asteroid is travelling directly towards the center of the Earth</u>, after impact ,it <u>does not impose any torque on earth's rotation,</u> So angular momentum of earth is conserved

⇒I_{1} \times W_{1} =I_{2} \times W_{2}

  • I_{1} is the moment of interia of earth before impact
  • W_{1} is the angular velocity of earth about an axis passing through the center of earth before impact
  • I_{2} is moment of interia of earth and asteroid system
  • W_{2} is the angular velocity of earth and asteroid system about the same axis

let  W_{1}=W

since \text{Time period of rotation}∝\frac{1}{\text{Angular velocity}}

⇒ if time period is to increase by 25%, which is \frac{5}{4} times, the angular velocity decreases 25% which is \frac{4}{5}  times

therefore W_{1} = \frac{4}{5} \times W_{1}

I_{1}=\frac{2}{5} \times M\times R^{2}(moment of inertia of solid sphere)

where M is mass of earth

           R is radius of earth

I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where M_{1} is mass of asteroid

⇒ \frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}+ M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})

\frac{1}{2} \times M\times R^{2}= (\frac{2}{5} \times M\times R^{2}+ M_{1}\times R^{2})

M_{1}\times R^{2}= \frac{1}{10} \times M\times R^{2}

⇒M_{1}=}\frac{1}{10} \times M

3 0
3 years ago
A 1000 kg truck moving at 2.0 m/s runs into a concrete wall. It takes 0.5 s for the truck to completely stop. What is the magnit
guajiro [1.7K]

Answer:

Momentum is given by

p

=

m

v

. Impulse is the change of momentum,

I

=

Δ

p

and is also equal to force times time:

I

=

F

t

. Rearranging,

F

=

I

t

=

Δ

p

t

=

0

−

20

,

000

5

=

−

4000

N

.

Explanation:

Momentum before the collision is

p

=

m

v

=

2000

⋅

10

=

20

,

000

k

g

m

s

−

1

.

Assuming the truck comes to a complete halt, the momentum after the collision is

0

k

g

m

s

−

1

.

The change in momentum,

Δ

p

, is initial minus final

→

0

−

20

,

000

=

−

20

,

000

This is called the impulse:

I

=

Δ

p

. Impulse is also equal (check the units) to force times time:

I

=

F

t

.

We can rearrange this expression to make

F

the subject:

F

=

I

t

=

Δ

p

t

=

−

20

,

000

5

=

−

4000

N

The negative sign just means the force acting is in the opposite direction to the initial momentum.

(This will be the average force acting during the collision: collisions are chaotic so the force is unlikely to be constant.)

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As denizens of the surface of a spinning planet, we are always in uniform circular motion. Imagine you are in Nairobi (on the Ea
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Answer:

6.02 radian

Explanation:

At 12 noon position is zero radian on Monday

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