Ask question

Ask question

All categories

3 years ago

5

8. In the billiard ball room at the basement of Memorial Union, two identical balls are traveling toward each other along a stra

ight line which is chosen to the x-axis. They experience an elastic head-on collision. Ignoring any frictional forces, if one ball m1 moving at a velocity of +4.67 m/s and the other m2 at a velocity of -7.89 m/s, what are the velocity (magnitude and direction) of each ball after the collision?

1 answer:

mel-nik [20]3 years ago

7 0

Answer:

- Ball 1 will move backward at 7.89 m/s

- Ball 2 will move forward at 4.67 m/s

Explanation:

In an elastic collision, both the total momentum and the total kinetic energy of the system are conserved.

The conservation of the momentum can be written as:

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where

$m_1$ is the mass of ball 1

$m_2$ is the mass of ball 2

$u_1$ is the initial velocity of ball 1

$u_2$ is the initial velocity of ball 2

$v_1$ is the final velocity of ball 1

$v_2$ is the final velocity of ball 2

The conservation of kinetic energy can be written as

$\frac{1}{2}m_1 u_2^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$

Working together the two equations, it is possible to find two expressions for the final velocities in terms of the initial velocities:

$v_1=\frac{m_1 -m_2}{m_1 +m_2}u_1+\frac{2m_2}{m_1+m_2}u_2\\v_2=\frac{2m_1}{m_1+m_2}u_1 -\frac{m_1 -m_2}{m_1 +m_2}u_2$

In this problem we have:

$m_1 = m_2 = m$ since the mass of the two balls is identical

$u_1=+4.67 m/s$ is the initial velocity of ball 1

$u_2=-7.89 m/s$ is the initial velocity of ball 2

Substituting into the equations, we find the final velocities:

$v_1=\frac{2m}{m+m}u_2=u_2 =-7.89 m/s\\v_2=\frac{2m}{m+m}u_1=u_1=+4.67 m/s$

Therefore:

- Ball 1 will move backward at 7.89 m/s

- Ball 2 will move forward at 4.67 m/s

You might be interested in

a 0.15 kg baseball has a momentum of 0.78 kg*m/s just before it lands on the ground. What was the ball's speed just before landi

vovangra [49]

5.2m/s

Explanation:

Given parameters:

Mass of baseball = 0.15kg

Momentum of baseball = 0.78kgm/s

Unknown:

Speed of baseball = ?

Solution:

The momentum of the baseball is a function of the product of the mass and velocity. It is a vector quantity:

Momentum = mass x velocity

Since the speed of the ball is unknown:

Velocity = $\frac{momentum }{mass}$

= $\frac{0.78}{0.15}$

= 5.2m/s

The speed of the baseball before it lands is 5.2m/s

Learn more:

Momentum brainly.com/question/9484203

#learnwithBrainly

4 0

3 years ago

What is the overall charge of 1.5 x 10^10 electrons?

Troyanec [42]

Answer: Charge = -2.4x10^-9 Coulombs

Explanation:

The charge of one electron is e = -1.6x10^-19 C

Then, the charge of 1.5 x 10^10 electrons is equal to 1.5 x 10^10 times the charge of one electron:

Here i will use the relation (a^b)*(a^c) = a^(b + c)

Charge = ( 1.5 x 10^10)*( -1.6x10^-19 C) = -2.4x10^(10 - 19) C

Charge = -2.4x10^-9 C

7 0

3 years ago

I need the answer for both of the questions please

Lady_Fox [76]

But even more pain on pain and then pain and pain ya feel me and even more pain okay and yes more pain

5 0

3 years ago

A rocket in deep space has an exhaust-gas speed of 2000 m/s. When the rocket is fully loaded, the mass of the fuel is five times

notka56 [123]

Answer:

$v_{f}$ = 1,386 m / s

Explanation:

Rocket propulsion is a moment process that described by the expression

$v_{f}$ - v₀ = $v_{r}$ ln (M₀ / Mf)

Where v are the velocities, final, initial and relative and M the masses

The data they give are the relative velocity (see = 2000 m / s) and the initial mass the mass of the loaded rocket (M₀ = 5Mf)

We consider that the rocket starts from rest (v₀ = 0)

At the time of burning half of the fuel the mass ratio is that the current mass is

M = 2.5 Mf

$v_{f}$ - 0 = 2000 ln (5Mf / 2.5 Mf) = 2000 ln 2

$v_{f}$ = 1,386 m / s

3 0

3 years ago

(a) What is the escape speed on a spherical asteroid whose radius is 500. km and whose gravitational acceleration at the surface

navik [9.2K]

Answer:

a) v= 1732.05m/s

b) d=250000m

c) v= 1414.214m/s

Explanation:

Notation

M= mass of the asteroid

m= mass of the particle moving upward

R= radius

v= escape speed

G= Universal constant

h= distance above the the surface

Part a

For this part we can use the principle of conservation of energy. for the begin the initial potential energy for the asteroid would be $U_i =-\frac{GMm}{R}$ .

The initial kinetic energy would be $\frac{1}{2}mv^2$ . The assumption here is that the particle escapes only if is infinetely far from the asteroid. And other assumption required is that the final potential and kinetic energy are both zero. Applying these we have:

$-\frac{GMm}{R}+\frac{1}{2}mv^2=0$ (1)

Dividing both sides by m and replacing $\frac{GM}{R}$ by $a_g R$

And the equation (1) becomes:

$-a_g R+\frac{1}{2} v^2=0$ (2)

If we solve for v we got this:

$v=\sqrt{2 a_g R}=\sqrt{2x3\frac{m}{s^2}x500000m}=1732.05m/s$

Part b

When we consider a particule at this surface at the starting point we have that:

$U_i=-\frac{GMm}{R}$

$K_i=\frac{1}{2}mv^2$

Considering that the particle is at a distance h above the surface and then stops we have that:

$U_f=-\frac{GMm}{R+h}$

$K_f=0$

And the balance of energy would be:

$-\frac{GMm}{R}+\frac{1}{2}mv^2 =-\frac{GMm}{R+h}$

Dividing again both sides by m and replacing $\frac{GM}{R}$ by $a_g R^2$ we got:

$-a_g R+\frac{1}{2}v^2 =-\frac{a_g R^2}{R+h}$

If we solve for h we can follow the following steps:

$R+h=-\frac{a_g R^2}{-a_g R+\frac{1}{2}v^2}$

And subtracting R on both sides and multiplying by 2 in the fraction part and reordering terms:

$h=\frac{2a_g R^2}{2a_g R-v^2}-R$

Replacing:

$h=\frac{2x3\frac{m}{s^2}(500000m)^2}{2(3\frac{m}{s^2})(500000m)-(1000m/s)^2}- 500000m=250000m$

Part c

For this part we assume that the particle is a distance h above the surface at the begin and start with 0 velocity so then:

$U_i=-\frac{GMm}{R+h}$

$K_i=0$

And after the particle reach the asteroid we have this:

$U_f=-\frac{GMm}{R}$

$K_f=\frac{1}{2}mv^2$

So the balance of energy would be:

$-\frac{GMm}{R+h}=-\frac{GMm}{R}+\frac{1}{2}mv^2$

Replacing again $a_g R^2$ instead of GM and dividing both sides by m we have:

$-\frac{a_g R^2}{R+h}=-a_g R+\frac{1}{2}v^2$

And solving for v:

$a_g R-\frac{a_g R^2}{R+h}=\frac{1}{2}v^2$

Multiplying both sides by two and taking square root:

$v=\sqrt{2a_g R-\frac{2a_g R^2}{R+h}}$

Replacing

$v=\sqrt{2(3\frac{m}{s^2})(500000m)-\frac{2(3\frac{m}{s^2}(500000m)^2}{500000+1000000m}}=1414.214m/s$

3 0

3 years ago