Question 16 of 20

A 100 g ball collides elastically with a 300 g ball that is at rest. If the 100 g ball was traveling

sammy [17]

Answer:

The magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).

Explanation:

We can find the velocity of the two balls after the collision by conservation of linear momentum and energy:

$P_{1} = P_{2}$

$m_{1}v_{1_{i}} + m_{2}v_{2_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}$

Where:

m₁: is the mass of the ball 1 = 100 g = 0.1 kg

m₂: is the mass of the ball 2 = 300 g = 0.3 kg

$v_{1_{i}}$ : is the initial velocity of the ball 1 = 6.20 m/s

$v_{2_{i}}$ : is the initial velocity of the ball 2 = 0 (it is at rest)

$v_{1_{f}}$ : is the final velocity of the ball 1 =?

$v_{2_{f}}$ : is the initial velocity of the ball 2 =?

$m_{1}v_{1_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}$

$v_{1_{f}} = v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}}$ (1)

Now, by conservation of kinetic energy (since they collide elastically):

$\frac{1}{2}m_{1}v_{1_{i}}^{2} = \frac{1}{2}m_{1}v_{1_{f}}^{2} + \frac{1}{2}m_{2}v_{2_{f}}^{2}$

$m_{1}v_{1_{i}}^{2} = m_{1}v_{1_{f}}^{2} + m_{2}v_{2_{f}}^{2}$ (2)

By entering equation (1) into (2) we have:

$m_{1}v_{1_{i}}^{2} = m_{1}(v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}})^{2} + m_{2}v_{2_{f}}^{2}$

$0.1 kg*(6.20 m/s)^{2} = 0.1 kg*(6.2 m/s - \frac{0.3 kg*v_{2_{f}}}{0.1 kg})^{2} + 0.3 kg(v_{2_{f}})^{2}$

By solving the above equation for $v_{2_{f}}$ :

$v_{2_{f}} = 3.1 m/s$

Now, $v_{1_{f}}$ can be calculated with equation (1):

$v_{1_{f}} = 6.20 m/s - \frac{0.3 kg*3.1 m/s}{0.1 kg} = -3.1 m/s$

The minus sign of $v_{1_{f}}$ means that the ball 1 (100g) is moving in the negative x-direction after the collision.

Therefore, the magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).

I hope it helps you!

5 0

3 years ago

What is the Isotope notation for a Lithium atom?

Serggg [28]

Lithium has five isotopes, and each will have different notation. Lithium-7 is commonly used in salt and nuclear reactors. .ZAX , where X is the atomic symbol for the element, here Li for lithium, Z is the mass number, or the total number of protons and neutrons, and A is the atomic number, or the number of protons

6 0

3 years ago

Kenitic Energy (KE)<br> 1.doubled

astra-53 [7]

Answer:

mass

Explanation:

This energy of motion is what we call kinetic energy. ... In fact, kinetic energy is directly proportional to mass: if you double the mass, then you double the kinetic energy. Second, the faster something is moving, the greater the force it is capable of exerting and the greater energy it possesses.

pls make as brainlieast

3 0

3 years ago

A rock is thrown upward from a bridge into a river below. The function f(t)=−16t2+44t+88 determines the height of the rock above

babymother [125]

1) 88 ft

2) 4.09 s

3) 1.38 s

4) 118.2 m

Explanation:

1)

For an object thrown upward and subjected to free fall, the height of the object at any time t is given by the suvat equation:

$h(t) = h_0 + ut - \frac{1}{2}gt^2$ (1)

where

$h_0$ is the height at time t = 0

$u$ is the initial vertical velocity

$g=32 ft/s^2$ is the acceleration due to gravity

The function that describes the height of the rock above the surface at a time t in this problem is

$f(t)=-16t^2+44t+88$ (2)

By comparing the terms with same degree of eq(1) and eq(2), we observe that

$h_0 = 88 ft$

which means that the rock is at height h = 88 ft when t = 0: therefore, this means that the height of the bridge above the water is 88 feet.

2)

The rock will hit the water when its height becomes zero, so when

$f(t)=0$

which means when

$0=-16t^2+44t+88$

First of all, we can simplify the equation by dividing each term by 4:

$0=-4t^2+11t+22$

This is a second-order equation, so we solve it using the usual formula and we find:

$t_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-11\pm \sqrt{(11)^2-4(-4)(22)}}{2(-4)}=\frac{-11\pm \sqrt{121+352}}{-8}=\frac{-11\pm 21.75}{-8}$

Which gives only one positive solution (we neglect the negative solution since it has no physical meaning):

t = 4.09 s

So, the rock hits the water after 4.09 seconds.

3)

Here we want to find how many seconds after being thrown does the rock reach its maximum height above the water.

For an object in free fall motion, the vertical velocity is given by the expression

$v=u-gt$

where

u is the initial velocity

g is the acceleration due to gravity

t is the time

The object reaches its maximum height when its velocity changes direction, so when the vertical velocity is zero:

$v=0$

which means

$0=u-gt$

Here we have

$u=+44 ft/s$ (initial velocity)

$g=32 ft/s^2$ (acceleration due to gravity)

Solving for t, we find the time at which this occurs:

$t=\frac{u}{g}=\frac{44}{32}=1.38 s$

4)

The maximum height of the rock can be calculated by evaluating f(t) at the time the rock reaches the maximum height, so when

t = 1.38 s

The expression that gives the height of the rock at time t is

$f(t)=-16t^2+44t+88$

Substituting t = 1.38 s, we find:

$f(1.38)=-16(1.38)^2 + 44(1.38)+88=118.2 m$

So, the maximum height reached by the rock during its motion is

$h_{max}=118.2 m$

Which means 118.2 m above the water.

3 0

4 years ago

First consider a Carnot engine that operates between a hot reservoir at 58°C and a cold reservoir at -17°C. In running 24 minute

jok3333 [9.3K]

Answer:

Explanation:

Given that,

Hot reservoir temperature is

TH = 58°C

TH = 58 + 273 = 331 K

Cold reservoir temperature

TC = -17°C

TC = -17°C + 273 = 256K

In 24minutes 590 J of heat was removed from hot reservoir

t = 24mins

t = 24 × 60 = 1440 seconds

Then,

Power is given as

P = E / t

P = 590 / 1440

P = 0.41 W

Since this removed from the hot reservoir, then, QH = 0.41W

We want to find heat expelled to the cold reservoir QX

Efficiency is given as

η = 1 - TH/TC = 1 - QH/QC

1 - TH/TC = 1 - QH/QC

TH/TC = QH/QC

Make QC subject of formula

QC = QH × TC / TH

QC = 0.41 × 256 / 331

QC = 0.317 W

7 0

3 years ago