Ask question

Ask question

All categories

3 years ago

5

When a 1.0-kilogram cart moving with a speed of 0.50 meter per second on a horizontal surface collides with a second 1.0-kilogra

m cart initially at rest, the carts lock together. What is the speed of the combined carts after the collision?

1 answer:

yanalaym [24]3 years ago

4 0

The speed of the combined carts after the collision is 0.25 m/s

Explanation:

We can solve this problem by using the principle of conservation of momentum. In fact, the total momentum of the system must be conserved before and after the collision, so we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v$

where:

$m_1 = 1.0 kg$ is the mass of the first cart

$u_1 = 0.50 m/s$ is the initial velocity of the first cart

$m_2 = 1.0 kg$ is the mass of the second cart

$u_2 = 0$ is the initial velocity of the second cart

$v$ is the final combined velocity of the two carts

Re-arranging the equation and substituting the values, we find: the final velocity:

$v=\frac{m_1 u_1}{m_1+m_2}=\frac{(1.0)(0.50)}{1.0+1.0}=0.25 m/s$

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

You might be interested in

An ice skater of mass m = 60 kg coasts at a speed of v = 0.8 m/s past a pole. At the distance of closest approach, her center of

IrinaVladis [17]

Answer:

1.78 rad/s

1.70344 rad/s

Explanation:

v = Velocity = 0.8 m/s

m = Mass of person = 60 kg

$r_1$ = Distance between center of mass of person and pole = 0.45 m

$r_2$ = New distance between center of mass of person and pole = 0.46 m

I = Moment of inertia

Angular speed is given by

$\omega_1=\dfrac{v}{r_1}\\\Rightarrow \omega_1=\dfrac{0.8}{0.45}\\\Rightarrow \omega_1=1.78\ rad/s$

The angular speed is 1.78 rad/s

In this system the angular momentum is conserved

$L_1=L_2\\\Rightarrow I_1\omega_1=I_2\omega_2\\\Rightarrow mr_1^2\omega_1=mr_2^2\omega_2\\\Rightarrow \omega_2=\dfrac{r_1^2\omega_1}{r_2^2}\\\Rightarrow \omega_2=\dfrac{0.45^2\times 1.78}{0.46^2}\\\Rightarrow \omega_2=1.70344\ rad/s$

The new angular speed is 1.70344 rad/s

7 0

3 years ago

The wheels of a wagon can be approximated as the combination of a thin outer hoop, of radius ????h=0.209 m and mass 4.32 kg , an

Stolb23 [73]

Answer:

r= 98.3 mm

Explanation:

For rim

R= 0.209 m

M= 4.32 kg

For rods

m= 7.37 kg

L= 2 R= 2 x 0.209 = 0.418 m

The Total moment of inertia of the wagon

I=MR²+2 x 1/12 m L²

Now by putting the values

$I=4.32\times 0.209^2+2\times \dfrac{1}{12}\times 7.73\times 0.418^2\ kg.m^2$

I=0.413 kg.m²

For disk:

t= 0.0462 m

Density ρ = 5990 kg/m³

Lets take r is the radius of disk

So the mass of the disc

m'=ρ πr² t

The moment of inertia of disc

I'=1/2 m'r²

I'=1/2 x r² x ρ πr² t

Given that

I = I'

1/2 x r² x ρ πr² t = 0.413 kg.m²

1/2 x r³ x ρ π t = 0.413

r³ x ρ π t = 0.826

$r^3=\dfrac{0.826}{\pi \times 5990 \times 0.0462}$

r³=0.00095

r=0.0983 m

r= 98.3 mm

8 0

3 years ago

Cheryl has a mug she says is made up of matter. Heather says that the hot choclate inside the cup is made up of matter too. Keat

dangina [55]

<u>Answer:</u>

Cheryl, Heather and Keaton all are correct.

<u>Explanation:</u>

Everything around you is made of matter and matter is anything that has mass and occupies space or in other words, anything which has volume is called matter.

Here, in the given example, Cheryl, Heather and Keaton all are correct because the mug, the hot chocolate which is inside the mug and the steam coming out of it, all have mass. Therefore, all are correct except for Mikayla.

6 0

3 years ago

Read 2 more answers

At what height h above the ground does the projectile have a speed of 0.5v?

maw [93]

Answer:

$h=\dfrac{3v^2}{8g}$

Explanation:

It is given that,

Speed of the projectile is 0.5 v. Let h is the height above the ground. Using the first equation of motion to find it.

$v=u+at$

$v=u-gt$

Initial speed of the projectile is v and final speed is 0.5 v.

$0.5v=v-gt$

$t=\dfrac{v}{2g}$

g is the acceleration due to gravity

Let h is the height above the ground. Using the second equation of motion as :

$h=vt-\dfrac{1}{2}gt^2$

$h=v\dfrac{v}{2g}-\dfrac{1}{2}g(\dfrac{v}{2g})^2$

$h=\dfrac{3v^2}{8g}$

So, the height of the projectile above the ground is $\dfrac{3v^2}{8g}$ . Hence, this is the required solution.

6 0

3 years ago

A block, M1=10kg, slides down a smooth, curved incline of height 5m. It collides elastically with another block, M2=5kg, which i

erma4kov [3.2K]

Answer:

2.86 m

Explanation:

Given:

M₁ = 10 kg

M₂ = 5 kg

$\mu_k$ = 0.5

height, h = 5 m

distance traveled, s = 2 m

spring constant, k = 250 N/m

now,

the initial velocity of the first block as it approaches the second block

u₁ = √(2 × g × h)

or

u₁ = √(2 × 9.8 × 5)

or

u₁ = 9.89 m/s

let the velocity of second ball be v₂

now from the conservation of momentum, we have

M₁ × u₁ = M₂ × v₂

on substituting the values, we get

10 × 9.89 = 5 × v₂

or

v₂ = 19.79 m/s

now,

let the velocity of mass 2 when it reaches the spring be v₃

from the work energy theorem, we have

Work done by the friction force = change in kinetic energy of the mass 2

or

$0.5\times5\times9.8\times2 = \frac{1}{2}\times5\times( v_3^2-19.79^2)$

or

v₃ = 20.27 m/s

now, let the spring is compressed by the distance 'x'

therefore, from the conservation of energy

we have

Energy of the spring = Kinetic energy of the mass 2

or

$\frac{1}{2}kx^2=\frac{1}{2}mv_3^2$

on substituting the values, we get

$\frac{1}{2}\times250\times x^2=\frac{1}{2}\times5\times20.27^2$

or

x = 2.86 m

8 0

3 years ago