Ask question

Ask question

All categories

2 years ago

8

Use the following information for question 26 and 27. A 550-g ball traveling at 8.0 m/s undergoes a sudden head-on perfectly ela

stic collision with a 250-g ball traveling toward it also at 8.0 m/s. What is the speed of the 250-g mass just after the collision

1 answer:

Gnoma [55]2 years ago

6 0

Answer:

The speed of 250 g ball after collision is 14 m/s.

Explanation:

mass of first ball, m = 550 g = 0.55 kg

initial velocity of first ball, u = 8 m/s

mass of second ball, m' = 250 g = 0.25 kg

initial velocity of second ball, u' = - 8 m/s

Let the speed of the balls is v and v' after the collision.

Use conservation of momentum

m u + m' u' = m v + m' v'

0.55 x 8 - 0.25 x 8 = 0.55 v + 0.25 v'

0.55 v + 0.25 v' = 2.4 ..... (1)

Use the formula of coefficient of restitution,

For elastic collision, e = 1

$e =\frac{v'-v}{u - u'}\\\\1 =\frac{v'-v}{8+8}\\\\16 =v' - v...... (2)$

By solving (1) and (2)

v = - 2 m/s and v' = 14 m/s

You might be interested in

__ is the second most abundant element in Earth’s crust. It is found in ___, and ___; and in _ which is used to make pottery.

lianna [129]

Answer:

silicon is the second most abundant element in Earth’s crust. It is found in the sun and stars; and in clay which is used to make pottery.

Explanation:

6 0

2 years ago

Which of the following is the closest to the scientific fact

Finger [1]

B - A theory seems to be the closest

7 0

2 years ago

Work is the transfer of _______ that occurs when a force makes an object move.

Vlad1618 [11]

I think it is energy

4 0

2 years ago

Three point charges are arranged on a line. Charge q3 = 5 nC and is at the origin. Charge q2 = - 3 nC and is at x = 4 cm. Charge

Taya2010 [7]

Answer:

q₁ = + 1.25 nC

Explanation:

Theory of electrical forces

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Known data

q₃=5 nC

q₂=- 3 nC

d₁₃= 2 cm

d₂₃ = 4 cm

Graphic attached

The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.

For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So, the charge q₁ must be positive(q₁+).

The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).

The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs. F₂₃ is directed to the right (+x)

Calculation of q1

F₁₃ = F₂₃

$\frac{k*q_{1}*q_3 }{(d_{13})^{2} } = \frac{k*q_{2}*q_3 }{(d_{23})^{2} }$

We divide by (k * q3) on both sides of the equation

$\frac{q_{1} }{(d_{13})^{2} } = \frac{q_{2} }{(d_{23})^{2} }$

$q_{1} = \frac{q_{2}*(d_{13})^{2} }{(d_{23} )^{2} }$

$q_{1} = \frac{5*(2)^{2} }{(4 )^{2} }$

q₁ = + 1.25 nC

3 0

2 years ago

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

vlada-n [284]

Answer:

$\frac{1}{10}$ M

Explanation:

To apply the concept of <u>angular momentum conservation</u>, there should be no external torque before and after

As the <u>asteroid is travelling directly towards the center of the Earth</u>, after impact ,it <u>does not impose any torque on earth's rotation,</u> So angular momentum of earth is conserved

⇒ $I_{1} \times W_{1} =I_{2} \times W_{2}$

$I_{1}$ is the moment of interia of earth before impact
$W_{1}$ is the angular velocity of earth about an axis passing through the center of earth before impact
$I_{2}$ is moment of interia of earth and asteroid system
$W_{2}$ is the angular velocity of earth and asteroid system about the same axis

let $W_{1}=W$

since $\text{Time period of rotation}∝$ $\frac{1}{\text{Angular velocity}}$

⇒ if time period is to increase by 25%, which is $\frac{5}{4}$ times, the angular velocity decreases 25% which is $\frac{4}{5}$ times

therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

$I_{1}=\frac{2}{5} \times M\times R^{2}$ (moment of inertia of solid sphere)

where M is mass of earth

R is radius of earth

$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid

⇒ $\frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}$ $+$ $M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})$

$\frac{1}{2} \times M\times R^{2}$ = $(\frac{2}{5} \times M\times R^{2}$ + $M_{1}\times R^{2})$

$M_{1}\times R^{2}=$ $\frac{1}{10} \times M\times R^{2}$

⇒ $M_{1}=}$ $\frac{1}{10} \times M$

3 0

2 years ago