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A 600gair-track glider collides with a spring at one end of the track. FIGURE EX11.12 shows the gliderโ€™s velocity and the force exerted on the glider by the spring. How long is the glider in contact with the spring?

Short Answer

Expert verified

The glider can takeโ–ณt=0.2sto in contact with the spring.

Step by step solution

01

Given information   

We need to find that How long is the glider in contact with the spring.

02

Simplify

We have to find โ–ณtbecause of the glider is in contact with spring at the time of impulse localid="1649920534377" โ–ณt.

For a short time a object receives a force is known as impulse. and under the area the curve of the force-versus-time graph and it is the same as momentum. The impulse is the quantity Jxand it is given by equation in the form

impulse=Jx=โˆซtitfFx(t)dt=areaundertheFx(t)curvebetweentiandtf(1)

In a graph of force versus time the force is in the range of time โ–ณt. the impulse takes the rectangle shape, where height is h=36Nand and the base is b=โ–ณt. The area of the rectangle is half the product of the heightand the base

A=12(height)(base)

Using equation (1)to get impulse

Jx=12(height)(base)=12(36N)โ–ณt=18โ–ณt

After finding impulse Jx=18โ–ณt.An impulse is the quantity Jxthat is delivered by a force to an object with an initial momentum pixthat changes its momentum pfx, and it is given by equation in the form of a force.

pfx=pix+Jx(2)

THe momentum is given by equation in the form

p=mv

03

Simplify 

By using the momentum pand impulse Jxexpressions in equation (2), we can find an expression for the time โ–ณtas follows

pfx=pix+Jxmvf=mvi+18โ–ณtโ–ณt=m(vf-vi)18(3)

In a graph of velocity versus time the initial speed is vi=-3m/sand final velocity vf=3m/s.

So puttin values into equation (3)for the mass m,viandvf to get โ–ณt

โ–ณt=m(vf-vi)18=0.60kg3.0m/s--3.0m/s18=0.2s

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