A cart with mass 340 g moving on a frictionless linear air track at an initial speed of 1.2 m/s undergoes an elastic collision with an initially stationary cart of unknown mass. After the collision, the first cart continues in its original direction at 0.66 m/s. (a) What is the mass of the second cart? (b) What is its speed after impact?

Answers

Answer 1
Answer:

Answer:

A) m2 = 98.71g

B) v_f2 = 1.86 m/s

Explanation:

We are given;

Mass of cart; m1 = 340g

Initial speed; v_i1 = 1.2 m/s

Final speed; v_f1 = 0.66 m/s

A)Since the collision is elastic, we can simply apply the conservation of momentum to get;

m1•(v_i1) = m1•(v_f1) + m2•(v_f2) - - - - - (eq1)

From conservation of kinetic energy, we have;

(1/2)m1•(v_i1)² = (1/2)m1•(v_f1)² + (1/2)m2•(v_f2)² - - - - eq(2)

Let's make v_f2 the subject in eq 2;

Thus,

v_f2 = √([m1•(v_i1)² - m1•(v_f1)²]/m2)

v_f2 = √([m1((v_i1)² - (v_f1)²)]/m2)

Let's put this for v_f2 in eq1 to obtain;

m2 = {m1((v_i1) - (v_f1))}/√([m1((v_i1)² - (v_f1)²)]/m2)

Let's square both sides to give;

(m2)² = {m1•m2((v_i1) - (v_f1))²}/([(v_i1)² - (v_f1)²]

This gives;

m2 = {m1((v_i1) - (v_f1))²}/([(v_i1)² - (v_f1)²]

Plugging in the relevant values to get;

m2 = {340((1.2) - (0.66))²}/([(1.2)² - (0.66)²]

m2 = 98.71g

B) from equation 1, we have;

m1•(v_i1) = m1•(v_f1) + m2•(v_f2)

Making v_f2 the subject, we have;

v_f2 = m1[(v_i1) - (v_f1)]/m2

Plugging in the relevant values to get;

v_f2 = 340[(1.2) - (0.66)]/98.71

v_f2 = 1.86 m/s

Answer 2
Answer:

Final answer:

To determine the mass of the second cart and its speed after impact, we can use the principle of conservation of momentum. The initial momentum of the first cart is equal to its final momentum plus the momentum of the second cart. After calculating the mass of the second cart, we can use the conservation of momentum again to find its speed by equating the final velocity of the combined carts to the initial velocity of the first cart.

Explanation:

To determine the mass of the second cart, we can use the principle of conservation of momentum. The initial momentum of the first cart, with a mass of 340 g and an initial velocity of 1.2 m/s, is equal to its final momentum plus the momentum of the second cart. Using this equation, we can solve for the mass of the second cart.


After calculating the mass of the second cart, we can use the conservation of momentum again to find its speed after the impact. Since the two carts stick together after the collision, the final velocity of the combined carts is equal to the initial velocity of the first cart. Using this equation, we can solve for the speed of the second cart.

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The electric field at the distance of 3.5 meters from an infinite wall of charges is 125 N/C. What is the magnitude of the electric field 1.5 meters from the wall?

Answers

Explanation:

It is given that,

Distance, r = 3.5 m

Electric field due to an infinite wall of charges, E = 125 N/C

We need to find the electric field 1.5 meters from the wall, r' = 1.5 m. Let it is equal to E'. For an infinite wall of charge the electric field is given by :

E=(\lambda)/(2\pi \epsilon_o r)

It is clear that the electric field is inversely proportional to the distance. So,

(E)/(E')=(r')/(r)

E'=(Er)/(r')

E'=(125* 3.5)/(1.5)  

E' = 291.67 N/C

So, the magnitude of the electric field 1.5 meters from the wall is 291.67 N/C. Hence, this is the required solution.

A 75-m-long train begins uniform acceleration from rest. The front of the train has a speed of 18 m/swhen it passes a railway worker who is standing 125 m from where the front of the train started. What will be the speed of the last car as it passes the worker?

Answers

Answer:22.76 m/s

Explanation:

Given

Train length(L)=75 m

Front of train after travelling 125 m is 18 m/s

Time taken by the front of train to cover 125 m

v^2-u^2=2as

18^2-0=2* a* 125

a=1.296 m/s^2

Speed of the last part of train when it passes the worker i.e. front of train has to travel has to travel  a distance of 125+75=200 m

v^2-u^2=2as

v^2=2* 1.296* 200

v=√(518.4)=22.76 m/s

A businessperson took a small airplane for a quick flight up the coast for a lunch meeting and then returned home. The plane flew a total of 4 hours, and each way the trip was 200 miles. What was the speed of the wind that affected the plane, which was flying at a speed of 120mph? Round your answer to the nearest whole number.

Answers

Answer:

Speed of the wind is 48.989 mph

Explanation:

We have given each trip is of 200 miles

So total distance = 200 +200 = 400 miles

Speed of the airplane = 120 mph

Let the speed of the wind = x mph

So the speed of the airplane with wind = 120+x

So time taken by airplane with wind = (200)/(120+x)

Speed of the airplane against the wind = 120 - x

So time taken by the airplane against the wind =(200)/(120-x)

Total time is given as t= 4 hour

So (200)/(120+x)+(200)/(120-x)=4

(200(120-x)+200(120+x))/((120+x)(120-x))=4

48000=57600-4x^2

4x^2=9600

x = 48.989 mph

Answer:

Explanation:

  Type                           Distance             Rate         Time

Headwind 200 120-r   200/120-r

Tailwind     200  120 - r  200/120 - r

We know the times add to 4, so we write the equation:

200/120−r +   200/120 + r = 4  

We multiply both sides by the LCD and simplify to get:

(120−r)(120+r) ((200/120 -r ) + 200/120+r) = 4(120 -r) (120 +r)

200(120−r)+200(120+r)=4(120−r)(120+r)

Factor the 200 and simplify inside the parentheses to find:

200(120−r+120+r)=4(1202−r2)

200(240)=4(1202−r2)

200(60)=120^2−r^2

12,000=14,400−r^2

−2,400= −r^2

49 ≈ r

The speed of the wind is 49mph.

A barbell spins around a pivot at its center at A. The barbell consists of two small balls, each with mass 450 grams (0.45 kg), at the ends of a very low mass rod of length d = 20 cm (0.2 m; the radius of rotation is 0.1 m). The barbell spins clockwise with angular speed 120 radians/s.What is the speed of ball 1?

Answers

The linear speed of the ball for the circular motion is determined as 12 m/s.

The given parameters;

  • mass of each ball, m = 450 g = 0.45 kg
  • length of the rod, L = 0.2 m
  • radius of the rod, r = 0.1 m
  • angular speed of the ball, ω = 120 rad/s

The linear speed of the ball is calculated as follows;

v = ωr

where;

  • ω is the angular speed of the ball
  • r is the radius of circular motion of the ball

The linear speed of the ball is calculated as follows;

v = ωr

v = 120 x 0.1

v = 12 m/s

Thus, the linear speed of the ball for the circular motion is determined as 12 m/s.

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Answer:

The speed of ball is 12 (m)/(s)

Explanation:

Given:

Mass of ball m = 0.45 kg

Radius of rotation r = 0.1 m

Angular speed \omega = 120 (rad)/(s)

Here barbell spins around a pivot at its center and barbell consists of two small balls,

From the formula of speed in terms of angular speed,

  v  = r \omega

Where v = speed of ball

  v = 120 * 0.1

  v = 12 (m)/(s)

Therefore, the speed of ball is 12 (m)/(s)

1. Two forces F~ 1 and F~ 2 are acting on a block of mass m=1.5 kg. The magnitude of force F~ 1 is 12N and it makes an angle of θ = 37◦ with the horizontal as shown in figure-1. The block is sliding at a constant velocity over a frictionless floor.(a) Find the value of the normal force on the block.

(b) Find the magnitude of force F~2 that is acting on the block

(c) Find the magnitude of force F~ 2 if the block accelerates with a magnitude of a = 2.5 m/s2 along the direction of F~ 2 .

Answers

Answer:

Normal force=7.48 N

Explanation:

N+F~1 sinθ-mg=0

=>N=1.5*9.8-12 sin37◦

=>N=14.7-7.22=7.48 N

A charge Q = 1.96 10-8 C is surrounded by an equipotential surface with a surface area of 1.18 m2. what is the electric potential at this surface?

Answers

Answer:

V = 575.6 Volts

Explanation:

As we know that surface area of the equi-potential surface is given as

A = 1.18 m^2

so we will say

A = 4\pi r^2

1.18 = 4\pi r^2

r = 0.31 m

Now the potential due to a point charge is given as

V = (kQ)/(r)

V = ((9* 10^9)(1.96 * 10^(-8)))/(0.31)

V = 575.6 Volts

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