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Cloud [144]
3 years ago
11

Two objects of equal mass collide on a horizontal frictionless surface. Before the collision, object A is at rest while object B

has a constant velocity of 12 m/s. After the collision, the two objects are stuck together. What is the speed of the composite body (A B) after the collision
Physics
1 answer:
fenix001 [56]3 years ago
7 0

Answer: 6m/s

Explanation:

Using the law of conservation of momentum, the change in momentum of the bodies before collision is equal to the change in momentum after collision.

After collision, the two objects will move at the same velocity (v).

Let mA and mB be the mass of the two objects

uA and uB be their velocities before collision.

v be their velocity after collision

Since the two objects has the same mass, mA= mB= m

Also since object A is at rest, its velocity = 0m/s

Velocity of object B = 12m/s

Mathematically,

mAuA + mBuB = (mA+mB )v

m(0) + m(12) = (m+m)v

0+12m = (2m)v

12m = 2mv

12 = 2v

v = 6m/s

Therefore the speed of the composite body (A B) after the collision is 6m/s

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<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2 /><h2>\huge\boxed{\text{V = 9.5 m/s}}</h2><h2>_____________________________________</h2>

<h2>DATA:</h2>

mass = m = 2kg

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TO FIND:

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<h2>SOLUTION:</h2>

According to the object of mass 2 kg travels a distance when the force was exerted on it. The graph between the Force and position was plotted which shows that 30 N of force was used to push the object till the distance of 6.0m.

To find the work, I will use the method of determining the area of the plotted graph. As the graph is plotted in the straight line between the Force and work, THE PICTURE ATTCHED SHOWS THE AREA COVERED IN BLUE AS WORK DONE AND HEIGHT AS 30m AND DISTANCE COVERED AS 6m To solve for the area(work) of triangle is given as,

{\Longrightarrow}\qquad \qquad \qquad W\ =\ \frac{1}{2}\;(Base)\:(Height)

Base is the x-axis of the graph which is Position i.e. 6m

Height is the y-axis of the graph which is Force i.e. 30N

So,

                           W\ =\ \frac{1}{2}\:6\:x\:30

                           W   =  90 J

The work done is 90 J.

According to the principle of work and kinetic energy (also known as the work-energy theorem) states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.

{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad K.E\\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ m\ V^2 \\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ 2\ (V_f-V_i)^2\\\\{V_i\ is\ 0\ because\ the\ object\ was\ initially\ at\ rest}\\\\ {\Longrightarrow}\qquad \qquad \qquad W\quad\ =\ \frac{1}{2}\ x\ 2\ (V_f-0)^2 \\\\{\Longrightarrow}\qquad \qquad \qquad 90\quad = \frac{1}{2}\ x\ 2\ (V_f)^2

\\\\{\Longrightarrow}\qquad \qquad \qquad V_f\quad =\ \sqrt{\frac{2\ (90)\ }{2}}\\\\{\Longrightarrow}\qquad \qquad \qquad \boxed {V_f\quad =\ 9.48\ m/s}

\boxed{The\ Velocity\ of\ the\ Object\ of\ mass\ 2kg\ at\ 6\ meters\ of\ distance\ was\ 9.48\ m/s}

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>

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