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Sonbull [250]
2 years ago
11

A body of mass 100g moving with a velocity of 10.0m/s collides with a wall .if after the collision it moves with a velocity of 2

.0m/s in the opposite direction,calculate the change in momentum
Physics
1 answer:
ch4aika [34]2 years ago
7 0

Answer:

-1.2 kg - m/s

Explanation:

\pink{\frak{Given}}\begin{cases}\textsf{ A body of mass 100g moving with a velocity of 10.0m/s collides with a wall .}\\\textsf{ After the collision it moves with a velocity of 2.0m/s in the opposite direction.}\end{cases}

And we need to find out the change in momentum of the body . Here ,

  • velocity before collision (u) = 10m/s
  • velocity after collision (v) = 2m/s .

We know that momentum is defined as amount of motion contained in a body . Mathematically ,

\sf\longrightarrow momentum (p)= mass(m) * velocity(v)

Therefore change in momentum will be,

\sf\longrightarrow \triangle p = mv - mu

Since the direction of velocity changes after the collision , the velocity will be -2m/s .

\sf\longrightarrow \Delta p = 100g( -2m/s -10m/s) \\

\sf\longrightarrow \Delta p =\dfrac{100}{1000}kg ( -12m/s)  \\

\sf\longrightarrow \Delta p   = 0.1 kg * -12m/s \\

\sf\longrightarrow \boxed{\bf \Delta p = -1.2 \ kg-m/s} \\

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Answer:

d=1.29*10^{-6}m

Explanation:

From the question we are told that:

Distance of wall from CD D=1.4

Second bright fringe y_2= 0.803 m

Let

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Generally the equation for Interference is mathematically given by

y=frac{n*\lambda*D}{d}

Where

d=\frac{n*\lambda*D}{y}

d=\frac{2*431 *10^{-9}m*1.4}{0.803}

d=1.29*10^{-6}m

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3 years ago
A smooth circular hoop with a radius of 0.800 m is placed flat on the floor. A 0.300-kg particle slides around the inside edge o
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Answer:

a)11.25 J

b)Number of revolution = 1

Explanation:

Given that

Radius ,r= 0.8 m

m= 0.3 kg

Initial speed ,u= 10 m/s

final speed ,v= 5 m/s

a)

Initial energy

KE_i=\dfrac{1}{2}mu^2

KE_i=\dfrac{1}{2}0.3\times 10^2

KEi= 15 J

Final kinetic energy

KE_f=\dfrac{1}{2}mv^2

KE_f=\dfrac{1}{2}0.3\times 5^2

KEf=3.75 J

The  energy transformed from mechanical to internal = 15 - 3.75 J = 11.25 J

b)

The minimum value to complete the circular arc

 V=\sqrt{r.g}

Now by putting the values

V=\sqrt{0.8\times 10}

V= 2.82 m/s

So kinetic energy KE

KE=\dfrac{1}{2}mV^2

KE=\dfrac{1}{2}0.3\times 2.82^2

KE=1.19 J

ΔKE= KEi - KE

ΔKE= 15- 1.19 J

ΔKE=13.80 J

The minimum energy required to complete 2 revolutions = 2 x 11.25 J

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Here 22.5 J is greater than 13.8 J.So the particle will complete only one revolution.

Number of revolution = 1

4 0
3 years ago
A car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver n
satela [25.4K]

Answer:

a) t1 = v0/a0

b) t2 = v0/a0

c) v0^2/a0

Explanation:

A)

How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0

Vf = 0

Vf = v0 - a0*t

0 = v0 - a0*t

a0*t = v0

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B)

How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.

at this point

U = 0

v0 = u + a0*t

v0 = 0 + a0*t

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C)

The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v0 again? Express the separation distance in terms of v0 and a0 . Your answer should be positive.

t1 = t2 = t

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and we know t = v0/a0

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Vf2 = v0^2- 2a0s1

2a0s1 = v0^2

s1 = v0^2 / 2a0

After the full stop;

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s2 = v0^2/2a0

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= (2v0^2)/a0 - v0^2/a0

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