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3 years ago

Which situations describe an elastic collision?

Physics

2 answers:

Charra [1.4K]3 years ago

4 0

Answer;

A. two glass marbles bounce off each other.

Explanation;

Elastic collisions are collisions in which both momentum and kinetic energy are conserved. The total system kinetic energy before the collision equals the total system kinetic energy after the collision. If total kinetic energy is not conserved, then the collision is referred to as an inelastic collision.

An elastic collision occurs when the two objects "bounce" apart when they collide. Two rubber balls are a good example. In an elastic collision, both momentum and kinetic energy are conserved. Almost no energy is lost to sound, heat, or deformation.

Elan Coil [88]3 years ago

3 0

Elastic collisions are collisions in which both momentum and kinetic energy are conserved. I think the correct answer from the choices listed above is option A. Two glass marbles bounce off each other is an example of an elastic collision. Hope this answers the question.

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Main function of gametes?

vichka [17]

Sexual reproduction. thats the answer i think

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2 years ago

If a student flicks a stationary 0.1 kg ball with 5N of force for 0.1 seconds. What is its final

Alisiya [41]

5m/s

Explanation:

Given parameters:

Mass of ball = 0.1kg

Force on the ball = 5N

time taken = 0.1s

Unknown:

final speed of the ball = ?

Solution:

According to newton's second law "the net force on a body is the product of its mass and acceleration".

Force = mass x acceleration equation 1

Acceleration =

V is the final velocity

U is the initial velocity

T is the time taken

U = O since it is a stationary body;

a = $\frac{V}{T}$

Input "a" into equation 1

F = m x $\frac{V}{T}$

5 = 0.1 x $\frac{V}{0.1}$

V = 5m/s

learn more:

Newton's laws brainly.com/question/11411375

#learnwithBrainly

3 0

3 years ago

Danielle exerts a 14.0 N force to compress a spring by a distance of 8.00 cm. What is the spring constant of this spring

Lady_Fox [76]

Answer:

175 N/m

Explanation:

Given:

Force = F= 14.0 N

Distance = x = 8.00 cm = 0.08 m

To find:

spring constant

Solution:

spring constant is calculated by using Hooke's law:

k = F/x

Putting the values in above formula:

k = 14.0 / 0.08

k = 175 N/m

4 0

2 years ago

(b) A ball is thrown from a point 1.50 m above the ground. The initial velocity is 19.5 m/s at

Nataly_w [17]

The answers are:

(i) 6.35m

(ii) 20.2 m/s

It seems like you already have the answer, but let me show you how to get it:

You have two givens:

Vi = 19.5m/s

Θ = 30°

dy = 1.50m (This is not the maximum height, just to be clear)

When working with these types of equations, you just need to know your kinematics equations. For projectiles launched at an angle, you will need to first break down the initial velocity (Vi) into its horizontal (x) and vertical (y) components.

*Now remember this, if you are solving for something in the horizontal movement always use only x-components. When solving for vertical movements, always use y-components.

Let's move on to breaking down the initial velocity into both y and x components.

Viy = SinΘVi = (Sin30°)(19.5m/s) = 9.75 m/s

Vix = CosΘVi = (Cos30°)(19.5m/s) = 16.89 m/s

Okay, so we have that down now. The next step is to decide which kinematics equation you will use. Because you have no time, you need to use the kinematic equation that is not time dependent.

(i) Maximum height above the ground

Remember that the object was thrown 1.50m above the ground. So we save that for later. First we need to solve for the maximum height above the horizontal, or the point where it was thrown.

The kinematics equation you will use is:

$Vf^{2} = Vi^{2}+2ad$

Where:

Vf = final velocity

Vi = inital velocity

a = 9.8m/s²

d = displacement

We will derive our displacement from this equation. And you will come up with this:

$d = \dfrac{Vf^{2}-Vi^{2}}{2a}$

Again, remember that we are looking for a vertical component or y-component because we are looking for HEIGHT. So we use this plugging in vertical values only.

Vf at maximum height is always 0m/s because at maximum height, objects stop. Also because gravity is a downwards force you will use -9.8m/s².

Vfy = 0 m/s a = -9.8m/s² Viy = 9.75m/s

$dy = \dfrac{Vfy^{2}-Viy^{2}}{2a}$

$dy = \dfrac{0^{2}-(9.75m/s)^{2}}{2(-9.8m/s^{2})}$

$dy = \dfrac{-95.0625m^{2}/s^{2}}{-19.6m/s^{2}}$

$dy = 4.85m$

So from the point it was thrown, it reached a height of 4.85m. Now we add that to the height it was thrown to get the MAXIMUM HEIGHT ABOVE THE GROUND.

4.85m + 1.50m = 6.35m

(ii) Speed before it strikes the ground. (Vf=resultant velocity)

Okay, so here we need to consider a couple of things. To get the VF we need to first figure out the final velocities of both the x and y components. We are combining them to get the resultant velocity.

Vfx = horizontal velocity = Initial horizontal velocity (Vix). This is because gravity is not acting upon the horizontal movement so it remains constant.

Vfy = ?

VF =?

We need to solve this, again, using the same formula, but this time, you need to consider we are moving downwards now. So this time, instead of Vfy being 0 m/s, Viy is now 0 m/s. This is because it started moving from rest.

$Vfy^{2} = Viy^{2}+2ad$