The study of motion is called

3. In a physics lab, 0.500-kg cart (Cart A) moving rightward with a speed of 100 m/s collides with a 1.50-kg cart (Cart B) movin

Alex787 [66]

Answer:

The speed of the two carts after the collision is 10 m/s.

Explanation:

Hi there!

The momentum of the system Cart A - Cart B is conserved because there is no external force acting on the system at the instant of the collision. Then, the momentum of the system before the collision will be equal to the momentum of the system after the collision. The momentum of the system is calculated as the sum of momenta of cart A and cart B:

initial momentum = mA · vA1 + mB · vB1

final momentum = (mA + mB) · vAB2

Where:

mA = mass of cart A = 0.500 kg

vA1 = velocity of cart A before the collision = 100 m/s

mB = mass of cart B = 1.50 kg.

vB1 = velocity of cart B before the collision = - 20 m/s

vAB2 = velocity of the carts that move as a single object = unknown.

(notice that we have considered leftward as negative direction)

Since the momentum of system remains constant:

initial momentum = final momentum

mA · vA1 + mB · vB1 = (mA + mB) · vAB2

Solving for vAB2:

(mA · vA1 + mB · vB1) / (mA + mB) = vAB2

(0.500 kg · 100 m/s - 1.50 kg · 20 m/s) / (0.500 kg + 1.50 kg) = vAB2

vAB2 = 10 m/s

The speed of the two carts after the collision is 10 m/s.

6 0

3 years ago

Train cars are coupled together by being bumped into one another. Suppose two loaded cars are moving toward one another, the fir

tia_tia [17]

Answer:

7560 Joules

Explanation:

$m_1$ = Mass of first car = $1.5\times 10^5\ kg$

$m_2$ = Mass of second car = $2\times 10^5\ kg$

$u_1$ = Initial Velocity of first car = 0.3 m/s

$u_2$ = Initial Velocity of second car = -0.12 m/s

v = Velocity of combined mass

As linear momentum of the system is conserved

$m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow v=\frac{m_1u_1 + m_2u_2}{m_1 + m_2}\\\Rightarrow v=\frac{1.5\times 10^5\times 0.3 + 2\times 10^5\times -0.12}{1.5\times 10^5 + 2\times 10^5}\\\Rightarrow v=0.06\ m/s$

Energy lost is

$\Delta E=\Delta E_i-\Delta E_f\\\Rightarrow \Delta=\frac{1}{2}(m_1u_1^2 + m_2u_2^2-(m_1+m_2)v^2)\\\Rightarrow \Delta=\frac{1}{2}(1.5\times 10^5\times 0.3^2 + 2\times 10^5\times (-0.12)^2-(1.5\times 10^5 + 2\times 10^5)\times 0.06^2)\\\Rightarrow \Delta=7560\ J$

The Energy lost in the collision is 7560 Joules

7 0

2 years ago

A spring is characterized by a spring constant of 60 N/m. How much potential energy does it store, when stretched by 1.0 cm?

Anestetic [448]

Answer:

3.0 x10^-3 J

Explanation:

The potential energy of a spring is given by PE = (0.5)k*x^2

Where

K: Spring Constant = 60 N/m

x: displacement of the spring from its equilibrium position = 1cm = 0.01m

Then PE = 0.5(60)(.01)^2 = 0.003J = 3.0 x10^-3 J

8 0

3 years ago

Read 2 more answers

A coin is dropped into a wishing well. It takes 1.1 seconds for a splash to be heard. Calculate the depth of the wishing well

Zigmanuir [339]

Answer:

If the wishing well was in a vacuum, then s=ut + 0.5 a t^2 (s=distance, ... wishing well if you drop a coin into it and hear the splash 10 seconds

Explanation:

8 0

2 years ago

A 4 kg textbook sits on a desk. It is pushed horizontally with a 50 N applied force against a 15 N frictional force.

GarryVolchara [31]

a) See free-body diagram in attachment

b) The book is stationary in the vertical direction

c) The net horizontal force is 35 N in the forward direction

d) The net force on the book is 35 N in the forward horizontal direction

e) The acceleration is $8.75 m/s^2$ in the forward direction

Explanation:

a)

The free-body diagram of a body represents all the forces acting on the body using arrows, where the length of each arrow is proportional to the magnitude of the force and points in the same direction.

From the diagram of this book, we see there are 4 forces acting on the book:

- The applied force, F = 50 N, pushing forward in the horizontal direction

- The frictional force, $F_f = 15 N$ , pulling backward in the horizontal direction (the frictional force always acts in the direction opposite to the motion)

- The weight of the book, $W=mg$ , where m is the mass of the book and $g=9.8 m/s^2$ is the acceleration of gravity, acting downward. We can calculate its magnitude using the mass of the book, m = 4 kg:

$W=(4)(9.8)=39.2 N$

- The normal reaction exerted by the desk on the book, N, acting upward, and balancing the weight of the book

b)

The book is in equilibrium in the vertical direction, therefore there is no motion.

In fact, the magnitude of the normal reaction (N) exerted by the desk on the book is exactly equal to the weight of the book (W), so the equation of motion along the vertical direction is

$N-W=ma$