1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
san4es73 [151]
4 years ago
14

A 0.140-kg glider is moving to the right with a speed of 0.80 m/s on a frictionless, horizontal air track. The glider has a head

-on collision with a 0.299-kg glider that is moving to the left with a speed of 2.28 m/s. Find the final velocity (magnitude and direction) of each glider if the collision is elastic.
Physics
2 answers:
Pavel [41]4 years ago
8 0

Answer:

The final velocity of the first glider is 3.39553 m/s in the opposite direction

The final velocity of the second glider is 0.31553 m/s in the same direction

Explanation:

m_1 = Mass of first glider = 0.14 kg

m_2 = Mass of second glider = 0.299 kg

u_1 = Initial Velocity of first glider = 0.8 m/s

u_2 = Initial Velocity of second glider = -2.28 m/s

v_1 = Final Velocity of first glider

v_2 = Final Velocity of second glider

As momentum and Energy is conserved

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

{\tfrac {1}{2}}m_{1}u_{1}^{2}+{\tfrac {1}{2}}m_{2}u_{2}^{2}={\tfrac {1}{2}}m_{1}v_{1}^{2}+{\tfrac {1}{2}}m_{2}v_{2}^{2}

From the two equations we get

v_{1}=\frac{m_1-m_2}{m_1+m_2}u_{1}+\frac{2m_2}{m_1+m_2}u_2\\\Rightarrow v_1=\frac{0.14-0.299}{0.14+0.299}\times 0.8+\frac{2\times 0.299}{0.14+0.299}\times -2.28\\\Rightarrow v_1=-3.39553\ m/s

The final velocity of the first glider is 3.39553 m/s in the opposite direction

v_{2}=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 0.14}{0.14+0.299}\times 0.8+\frac{0.299-0.14}{0.14+0.299}\times -2.28\\\Rightarrow v_2=-0.31553\ m/s

The final velocity of the second glider is 0.31553 m/s in the same direction

liq [111]4 years ago
8 0

Answer:

v1 = 2.76 m/s and v2 = - 0.32 m/s

Explanation:

m1 = 0.140 kg

m2 = 0.299 kg

u1 = 0.80 m/s

u2 = - 2.28 m/s

Let the speed after collision is v1 and v2.

Use conservation of momentum

m1 x u1 + m2 x u2 = m1 x v1 + m2 x v2

0.140 x 0.80 - 0.299 x 2.28 = 0.140 x v1 + 0.299 x v2

0.112 - 0.68 = 0.14 v1 + 0.299 v2

0.14 v1 + 0.299 v2 = - 0.568 ..... (1)

By the use of coefficient of restitution, the value of e = 1 for elastic collision

e=\frac{v_{1}-v_{2}}{u_{2}-u_{1}}

u2 - u1 = v1 - v2

- 2.28 - 0.8 = v1 - v2

v1 - v2 = 3.08

v1 = 3.08 + v2

Put in equation (1)

0.14 (3.08 + v2) + 0.299 v2 = - 0.568

0.43 + 0.44 v2 = - 0.568

v2 = - 0.32 m/s

and

v1 = 3.08 - 0.32 = 2.76 m/s

Thus, v1 = 2.76 m/s and v2 = - 0.32 m/s

You might be interested in
FAST PLEASE What is the change in internal energy if 90 J of thermal energy is added to a
vladimir1956 [14]
A is right because I took the test
5 0
2 years ago
A cannon with a muzzle velocity of 500. meters per second fires a cannonball at an angle of 30.° above the horizontal. What is t
Reil [10]

Complete question

A cannon with a muzzle velocity of 500. meters per second fires a cannonball at an angle of 30.° above the horizontal. What is the vertical component of the cannonball's velocity as it leaves the cannon?

A   0.0 m/s

B    250. m/s

C   433 m/s

D    500. m/s

Answer:

The correct option is  C

Explanation:

From the question we are told that

   The velocity is  v  = 500 \ m/s

    The angle is  \theta = 30^o

   

Generally the vertical component of the canon ball is mathematically represented as  

          v_y = v * cos (30 )

=>       v_y = 500 * cos (30 )

=>       v_y = 433 \ m/s

7 0
3 years ago
Read 2 more answers
In a supermarket, you place a 22.3-N (around 5 lb) bag of oranges on a scale, and the scale starts to oscillate at 2.7 Hz. What
allsm [11]

Answer:

Force constant, k = 653.3 N/m

Explanation:

It is given that,

Weight of the bag of oranges on a scale, W = 22.3 N

Let m is the mass of the bag of oranges,

m=\dfrac{W}{g}

m=\dfrac{22.3}{9.8}

m = 2.27 kg

Frequency of the oscillation of the scale, f = 2.7 Hz

We need to find the force constant (spring constant) of the spring of the scale. We know that the formula of the frequency of oscillation of the spring is given by :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

k=4\pi^2 f^2m

k=4\pi^2 \times (2.7)^2\times 2.27

k = 653.3 N/m

So, the force constant of the spring of the scale is 653.3 N/m. Hence, this is the required solution.

7 0
4 years ago
Gary saved dimes and nickels at a ratio of 5:7. Then, he saved 20% more dimes and 3 times the original number of nickels. If he
TiliK225 [7]

Answer:

i think 53 or 56

Explanation:

cause i cross multiplied and got that

8 0
3 years ago
Read 2 more answers
Scenario
Anvisha [2.4K]

Answer:

1) t = 23.26 s,  x = 8527 m, 2)   t = 97.145 s,  v₀ = 6.4 m / s

Explanation:

1) First Scenario.

After reading your extensive problem, we are going to solve it, for this exercise we must use the parabolic motion relationships. Let's carry out an analysis of the situation, for deliveries the planes fly horizontally and we assume that the wind speed is zero or very small.

Before starting, let's reduce the magnitudes to the SI system

         v₀ = 250 miles/h (5280 ft / 1 mile) (1h / 3600s) = 366.67 ft/s

         y = 2650 m

Let's start by looking for the time it takes for the load to reach the ground.

         y = y₀ + v_{oy} t - ½ g t²

in this case when it reaches the ground its height is zero and as the plane flies horizontally the vertical speed is zero

         0 = y₀ + 0 - ½ g t2

          t = \sqrt{ \frac{2y_o}{g} }

          t = √(2 2650/9.8)

          t = 23.26 s

this is the horizontal scrolling time

          x = v₀ t

          x = 366.67  23.26

          x = 8527 m

the speed at the point of arrival is

         v_y = v_{oy} - g t = 0 - gt

         v_y = - 9.8 23.26

         v_y = -227.95 m / s

Module and angle form

        v = \sqrt{v_x^2 + v_y^2}

         v = √(366.67² + 227.95²)

        v = 431.75 m / s

         θ = tan⁻¹ (v_y / vₓ)

         θ = tan⁻¹ (227.95 / 366.67)

         θ = - 31.97º

measured clockwise from x axis

We see that there must be a mechanism to reduce this speed and the merchandise is not damaged.

2) second scenario. A catapult located at the position x₀ = -400m y₀ = -50m with a launch angle of θ = 50º

we look for the components of speed

           cos θ = v₀ₓ / v₀

           sin θ = v_{oy} / v₀

            v₀ₓ = v₀ cos θ

            v_{oy} = v₀ sin θ

we look for the time for the arrival point that has coordinates x = 0, y = 0

            y = y₀ + v_{oy} t - ½ g t²

            0 = y₀ + vo sin θ t - ½ g t²

            0 = -50 + vo sin 50 t - ½ 9.8 t²

            x = x₀ + v₀ₓ t

            0 = x₀ + vo cos θ t

            0 = -400 + vo cos 50 t

podemos ver que tenemos un sistema de dos ecuación con dos incógnitas

          50 = 0,766 vo t – 4,9 t²

          400 =   0,643 vo t

resolved

          50 = 0,766 ( \frac{400}{0.643 \ t}) t – 4,9 t²

          50 = 476,52 t – 4,9 t²

          t² – 97,25 t + 10,2 = 0

we solve the quadratic equation

         t = [97.25 ± \sqrt{97.25^2 - 4 \ 10.2}] / 2

         t = 97.25 ±97.04] 2

         t₁ = 97.145 s

         t₂ = 0.1 s≈0

the correct time is t1 the other time is the time to the launch point,

         t = 97.145 s

let's find the initial velocity

         x = x₀ + v₀ cos 50 t

         0 = -400 + v₀ cos 50 97.145

         v₀ = 400 / 62.44

         v₀ = 6.4 m / s

5 0
3 years ago
Other questions:
  • Is the 3 in the molecule a coefficient, subscript, or element? 3H₂O₂
    10·1 answer
  • How are earthquakes distributed on the map
    13·1 answer
  • What is the diffrence between a primary wave and a surface wave?
    13·1 answer
  • The medium seismic waves travel through is
    7·2 answers
  • As the size of the nucleus increases, more______are needed to maintain the attractive force.
    7·2 answers
  • OMG BRAINLIEST ANSWER AND 13POINTS OMG!!!!!!!!! THIS IS A ONE TIME OFFER FOLKS SO COME IN NOW!!FOLKS!!
    7·1 answer
  • A cheetah has an acceleration rate of 5.80 m/s2 . What is the cheetah’s final velocity after 7.80 seconds, if initially, his vel
    11·1 answer
  • States of matter question
    9·1 answer
  • A roller coaster is moving 10 m/s and a mass of 7 kg. The roller coaster has
    9·1 answer
  • Consider an electromagnetic wave propagating through a region of empty space. How is the energy density of the wave partitioned
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!