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3 years ago

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Two identical bodies are sliding toward each other on a frictionless surface. One moves at 1 m/s and the other at 2 m/s. They co

llide and stick. The speed of the combined mass is

1 answer:

mario62 [17]3 years ago

6 0

Answer:

$V=\dfrac{3}{2}\ m/s$

Explanation:

Two identical bodies are sliding toward each other on a frictionless surface.

Initial speed of body 1, m₁ = 1 m/s

Initial speed of body 2, m₂ = 2 m/s

They collide and stick.

We need to find the speed of the combined mass. Let V is the speed of the combined mass.

Using the conservation of momentum.

$m_1u_1+m_2u_2=V(m_1+m_2)$

We have, m₁ = m₂ = m

$m\times 1+m\times 2=(m+m)V\\\\m+2m=2m\times V\\\\3m=2mV\\\\V=\dfrac{3}{2}\ m/s$

So, the speed of the combined mass is $\dfrac{3}{2}\ m/s$ .

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Strontium. Hope that helps

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When the two surfaces of a thin film are parallel and the light is incident at some angle what fringes are formed? A. Circular B

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Answer:

The correct option is;

A. Circular

Explanation:

Some of the light that impinges on the surface are reflected and the rest are transmitted to a different medium

At the surface of the next medium also, some of the light are transmitted while the others are reflected and refracted through the first medium

The speed of light (and hence the wavelength and color) refracted through the thin film is changed as the distance the refracted light travels through the thin film is increased as we move away from the point directly in the front view to some distance as the reflected light path from those distance to the eye is increased due to their inclination giving them a different wavelength which are all equal at a radial distance from the eye hence forming a circular fringes.

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3 years ago

Two cars, A and B , travel in a straight line. The distance of A from the starting point is given as a function of time by xA(t)

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A) Car A is initially ahead

B) The two cars are at the same point at the times: t = 0, t = 2.27 s and

t = 5.73 s

C) The distance between the two cars is not changing at t = 1.00 s and t = 4.33 s

D) The two cars have same acceleration at t = 2.67 s

Explanation:

A)

The position of the two cars at time t is given by the following functions:

$x_A(t) = \alpha t + \beta t^2$

with

$\alpha = 2.60 m/s\\\beta = 1.20 m/s^2$

Substituting,

$x_A(t)=2.60t+1.20 t^2$

And

$x_B(t)=\gamma t^2 - \delta t^3$

with

$\gamma=2.80 m/s^2\\\delta = 0.20 m/s^3$

Substituting,

$x_B(t)=2.80t^2-0.20t^3$

Here we want to find which car is ahead just after they leave the starting point. To find that, we just need to calculate the position of the two cars after a very short amount of time, let's say at t = 0.1 s. Substituting this value into the two equations, we get:

$x_A(0.1)=2.60(0.1)+1.20(0.1)^2=0.27 m$

$x_B(0.1)=2.80(0.1)^2-0.20(0.1)^3=0.03 m$

So, car A is initially ahead.

B)

The two cars are at the same point when their position is the same. Therefore, when

$x_A(t)=x_B(t)$

which means when

$2.60t+1.20t^2 = 2.80t^2-0.20t^3$

Re-arranging the equation, we find

$0.20t^3-1.6t^2+2.60t=0\\t(0.20t^2-1.6t+2.60)=0$

One solution of this equation is t = 0 (initial point), while we have two more solutions given by the equation

$0.20t^2-1.6t+2.60=0$

which has two solutions:

t = 2.27 s

t = 5.73 s

So, these are the times at which the cars are at the same point.

C)

The distance between the two cars A and B is not changing when the velocities of the two cars is the same.

The velocity of car A is given by the derivative of the position of car A:

$v_A(t) = x_A'(t)=(2.60t+1.20t^2)'=2.60+2.40t$

The velocity of car B is given by the derivative of the position of car B:

$v_B(t)=x_B'(t)=(2.80t^2-0.20t^3)'=5.60t-0.60t^2$

Therefore, the distance between the two cars is not changing when the two velocities are equal:

$v_A(t)=v_B(t)\\2.60+2.40t=5.60t-0.60t^2\\0.60t^2-3.20t+2.60=0$

This is another second-order equation, which has two solutions:

t = 1.00 s

t = 4.33 s

D)

The acceleration of each car is given by the derivative of the velocity of the car A.

The acceleration of car A is:

$a_A(t)=v_A'(t)=(2.60+2.40t)'=2.40$

While the acceleration of car B is:

$a_B(t)=v_B'(t)=(5.60t-0.60t^2)'=5.60-1.20t$

So, the two cars have same acceleration when

$a_A(t)=a_B(t)$

And solving the equation, we find:

$2.40=5.60-1.20t\\1.20t=3.20\\t=2.67 s$

So, the two cars have same acceleration at t = 2.67 s.

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