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3 years ago

(a) Write the energy equation for an elastic collision. (b) For an inelastic collision.

Physics

1 answer:

velikii [3]3 years ago

5 0

Answer:

Explanation:

There are two types of collision.

(a) Elastic collision: When there is no loss of energy during the collision, then the collision is said to be elastic collision.

In case of elastic collision, the momentum is conserved, the kinetic energy is conserved and all the forces are conservative in nature.

The momentum of the system before collision = the momentum of system after collision

The kinetic energy of the system before collision = the kinetic energy after the collision

(b) Inelastic collision: When there is some loss of energy during the collision, then the collision is said to be inelastic collision.

In case of inelastic collision, the momentum is conserved, the kinetic energy is not conserved, the total mechanical energy is conserved and all the forces or some of the forces are non conservative in nature.

The momentum of the system before collision = the momentum of system after collision

The total mechanical energy of the system before collision = total mechanical of the system after the collision

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A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro

svetoff [14.1K]

Answer:

The equation of motion is $x(t)=-$ $\frac{1}{3} cos4\sqrt{6t}$

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is $32ft/s^2$

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet $x=\frac{4}{12} =\frac{1}{3}feet$

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

$W=kx$ ⇒ $k=\frac{W}{x}$

The spring constant , $k=\frac{24}{1/3}$

= 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

$m\frac{d^2x}{dt} +kx=0$

$\frac{3}{4} \frac{d^2x}{dt} +72x=0$

$\frac{d^2x}{dt} +96x=0$

Auxiliary equation is, $m^2+96=0$

$m=\sqrt{-96}$

= $\frac{+}{} i4\sqrt{6}$

Thus , the solution is $x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}$

$x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6t}$

The mass is released from the rest x'(0) = 0

$=-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6(0)}$ =0

$c_2$ $4\sqrt{6} =0$

$c_2=0$

Therefore , $x(t)=c_1$ $cos 4\sqrt{6t}$

Since , the mass is released from the rest from 4 inches

$x(0)= -4$ inches

$c_1 cos 4\sqrt{6(0)} =-\frac{4}{12}$ feet

$c_1=-\frac{1}{3}$ feet

Therefore , the equation of motion is $-\frac{1}{3} cos4\sqrt{6t}$

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Answer:

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Explanation:

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In attempting to pull a 1500-kg car out of a ditch, Arnold exerts a force of 300 N for three seconds, Bob exerts a force of 500

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Answer:

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Given:

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Impulse = Force × TIme

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Answer:

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Explanation:

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