Ask question

Ask question

All categories

2 years ago

6

A small, 150 g cart is moving at 1.60 m/s on a frictionless track when it collides with a larger, 5.00 kg cart at rest. After th

e collision, the small cart recoils at 0.890 m/s. What is the speed of the large cart after the collision?

1 answer:

NemiM [27]2 years ago

5 0

Answer:

Mass of larger cart will be 0.0213 m /sec

Explanation:

We have given mass of small cart $m_1=150gram=0.150kg$

Initial velocity of small cart $v_{1i}=1.6m/sec$

Mass of larger block $m_2=5kg$

Initial velocity of larger cart $v_{2i}=0m/sec$

After the collision velocity of smaller cart $v_{1f}=-0.890m/sec$

Now according conservation of momentum

$m_1v_{1i}+m_2v_{2i}=m_1v{1f}+m_2v_{2f}$

$0.150\times 1.6+5\times 0=0.150\times -0.890+5\times v_{2f}$

$5\times v_{2f}=0.1065$

$v_{2f}=0.0213m/sec$

You might be interested in

Corey runs a 100-meter race. 7 seconds after the race started Corey is 45 meters from the starting line and reaches his max spee

RUDIKE [14]

Answer:

Corey's max speed is $7 \frac{m}{s}$
the distance Corey's covers in z seconds is $7 \frac{m}{s} * z \ s$
$d (z) = 45 m + 7 \frac{m}{s} * z$
$d (x) = 45 m + 7 \frac{m}{s} * (x-7 s)$

Explanation:

<h3>Corey's max speed</h3>

For constant speed, we know:

$v=\frac{distance}{time}$

The distance between the 80 meters and the 45 meters is:

$distance = 80 m - 45 m = 35 m$

and the time it took to reach the 80 meter will be:

$time = 12 s - 7 s = 5 s$

So, Corey's max speed is

$v_{max}=\frac{35 m}{5 s} = 7 \frac{m}{s}$

<h3>How far runs Corey</h3>

As the velocity of Corey's is $v_{max}$ , the distance Corey's covers in z seconds is

$distance = v_{max} * z \ s$

$distance = 7 \frac{m}{s} * z \ s$

<h3>What is Corey's distance from the starting line</h3>

At time 7 + z seconds the distance will be the 45 meters he covers in the first part of the race plus the distance he traveled at constant speed. this is:

$d (z) = 45 m + v_{max} * z$

$d (z) = 45 m +7 \frac{m}{s} * z$

At time x ( x greater or equal to 7 seconds) the distance will be the 45 meters he covers in the first part of the race plus the distance he traveled at constant speed. this is:

$d (x) = 45 m + v_{max} * (x-7 s)$

$d (x) = 45 m + 7 \frac{m}{s} * (x-7 s)$

4 0

2 years ago

The fictional rocket ship Adventure is measured to be 50 m long by the ship's captain inside the rocket.When the rocket moves pa

tensa zangetsu [6.8K]

Complete question:

Part A:) The fictional rocket ship Adventure is measured to be 50 m long by the ship's captain inside the rocket.When the rocket moves past a space dock at 0.5c , space-dock personnel measure the rocket ship to be 43.3 m long. The rocket ship Adventure travels to a star many light-years away, then turns around and returns at the same speed. When it returns to the space dock, who would have aged less: the space-dock personnel or ship's captain?

Part B: What is the momentum of a proton traveling at 0.62 c ?

Answer

a)Who would have aged less=The Captain would have aged less

b) $p=3.96*10^{-19}kgm/s$

Explanation:

From the question we are told that

Length measured by captain $l_c=50m$

Speed of rocket past tje space dock $V=0.5c$

Length measured by space-dock personnel $l_c=43.3m$

a)

Generally time moves slower when moving at speed of light, due to time dilation or variation.

Who would have aged less=The Captain would have aged less

b)

Generally the equation for Relativistic Momentum is mathematically given as

$p=\frac{m*v}{1 - v^2/c^2}$

$p=\frac{1.67*10^2-27*0.62*3.0*10^8)}{\sqrt{ 1 -0.6^2}}$

$p=3.96*10^{-19}kgm/s$

3 0

2 years ago

a charge of 30. coulombs passes through a 24-ohm resistor in 6.0 seconds. what is the current through the resistor? (1) 1.3 a (3

DochEvi [55]

As I = Q/t (charge/time) the answer is 5a.

30/6 = 5

7 0

2 years ago

Read 2 more answers

A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.

ivann1987 [24]

Answer:

v₂ = 176.24 m/s

Explanation:

given,

angle of projectile = 45°

speed = v₁ = 150 m/s

for second trail

speed = v₂ = ?

angle of projectile = 37°

maximum height attained formula,

$H_{max}= \dfrac{v^2 sin^2(\theta)}{g}$

now,

$H_{max}= \dfrac{v_1^2 sin^2(\theta_1)}{g}$

$H_{max}= \dfrac{v_2^2 sin^2(\theta_2)}{g}$

now, equating both the equations

$\dfrac{v_2^2}{v_1^2}=\dfrac{sin^2(\theta_1)}{sin^2(\theta_2)}$

$\dfrac{v_2^2}{150^2}=\dfrac{sin^2(45^0)}{sin^2(37^0)}$

v₂² = 31061.79

v₂ = 176.24 m/s

velocity of projectile would be equal to v₂ = 176.24 m/s

8 0

2 years ago

what was the acceleration of the cart with Low fan speed cm/s squared? what was the acceleration of the cart with medium fan spe

ArbitrLikvidat [17]

Explanation:

The attached figure shows data for the cart speed, distance and time.

For low fan speed,

Distance, d = 500 cm

Time, t = 7.4 s

Average velocity,

$v=\dfrac{d}{t}\\\\v=\dfrac{500}{7.4}\\\\v=67.56\ cm/s$

Acceleration,

$a=\dfrac{v}{t}\\\\a=\dfrac{67.56}{7.4}\\\\a=9.12\ cm/s^2$

For medium fan speed,

Distance, d = 500 cm

Time, t = 6.4 s

Average velocity,

$v=\dfrac{d}{t}\\\\v=\dfrac{500}{6.4}\\\\v=78.12\ cm/s$

Acceleration,

$a=\dfrac{v}{t}\\\\a=\dfrac{78.12}{6.4}\\\\a=12.2\ cm/s^2$

For high fan speed,

Distance, d = 500 cm

Time, t = 5.6 s

Average velocity,

$v=\dfrac{d}{t}\\\\v=\dfrac{500}{5.6}\\\\v=89.28\ cm/s$

Acceleration,

$a=\dfrac{v}{t}\\\\a=\dfrac{89.28}{5.6}\\\\a=15.94\ cm/s^2$

Hence, this is the required solution.

8 0

2 years ago

Read 2 more answers