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kvv77 [185]
4 years ago
13

A striped billiard ball moves toward the right with speed 3 m/s. A solid billiard ball with the same mass moves toward the left

at the same speed. The two balls collide head on, causing each ball to reverse its direction of motion. If the collision is elastic, what is the final speed of the striped ball in m/s
Physics
1 answer:
Helen [10]4 years ago
4 0

Answer:

Final speed of striped ball is 3 m/s in left direction .

Explanation:

Given :

Two billiard ball with the same mass moves toward the left at the same speed 3 m/s .

Let , us assume right hand side direction to be positive and left hand side direction to be negative .

Also , let speed of ball after collision is (striped ball ) u and (solid ball) v .

It is also given that the collision is elastic .

Therefore , kinetic energy is conserved .

\dfrac{m(3)^2}{2}+\dfrac{m(3)^2}{2}=\dfrac{mu^2}{2}+\dfrac{mv^2}{2}\\\\u^2+v^2=18 ...... ( 1 )

Also , by conserving linear momentum .

We get :

3m-3m=mu+mv\\u=-v                ...... ( 2 )

Putting value of u from equation 2 to equation 1 .

We get :

2v^2=18\\v=3\ m/s

And , u = -3 m/s .

Therefore , final speed of striped ball is 3 m/s in left direction .

Hence , this is the required solution .

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kolezko [41]

Answer:

The fraction of kinetic energy lost in the collision in term of the initial energy is 0.49.

Explanation:

As the final and initial velocities are known it is possible then the kinetic energy is possible to calculate for each instant.

By definition, the kinetic energy is:

k = 0.5*mV^2

Expressing the initial and final kinetic energy for cars A and B:

ki=0.5*maVa_{i}^2+0.5*mbVb_{i}^2

kf=0.5*maVa_{f}^2+0.5*mbVb_{f}^2

Since the masses are equals:

m=ma=mb

For the known velocities, the kinetics energies result:

ki=0.5*mVa_{i}^2

ki=0.5*m(35 m/s)^2=612.5m^2/s^2*m

kf=0.5*mbVb_{f}^2

kf=0.5*m(25 m/s)^2=312.5m^2/s^2*m

The lost energy in the collision is the difference between the initial and final kinectic energies:

kl=ki-kf

kl = 612.5m^2/s^2*m-312.5 m^2/s^2*m=300 m^2/s^2*m

Finally the relation between the lost and the initial kinetic energy:

kl/ki = 300 m^2/s^2 * m / 612.5 m^2/s^2 * m

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7 0
4 years ago
A 1.8 kg uniform rod with a length of 90 cm is attached at one end to a frictionless pivot. It is free to rotate about the pivot
Leokris [45]

Answer:

a. 32.67 rad/s²  b. 29.4 m/s²

Explanation:

a. The initial angular acceleration of the rod

Since torque τ = Iα = WL (since the weight of the rod W is the only force acting on the rod , so it gives it a torque, τ at distance L from the pivot )where I = rotational inertia of uniform rod about pivot = mL²/3 (moment of inertia about an axis through one end of the rod), α = initial angular acceleration, W = weight of rod = mg where m = mass of rod = 1.8 kg and g = acceleration due to gravity = 9.8 m/s² and L = length of rod = 90 cm = 0.9 m.

So, Iα = WL

mL²α/3 = mgL

dividing through by mL, we have

Lα/3 = g

multiplying both sides by 3, we have

Lα = 3g

dividing both sides by L, we have

α = 3g/L

Substituting the values of the variables, we have

α = 3g/L

= 3 × 9.8 m/s²/0.9 m

= 29.4/0.9 rad/s²

= 32.67 rad/s²

b. The initial linear acceleration of the right end of the rod?

The linear acceleration at the initial point is tangential, so a = Lα = 0.9 m × 32.67 rad/s² = 29.4 m/s²

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