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Novosadov [1.4K]
3 years ago
8

A 4 kg body is traveling at 3 m/s with no external force acting on it. At a certain instant an internal explosion occurs, splitt

ing the body into two chunks of 2 kg mass each. The explosion gives the chunks an additional 18 J of kinetic energy. Neither chunk leaves the line of original motion. Determine the speed and direction of motion of each of the chunks after the explosion.
Physics
1 answer:
Ann [662]3 years ago
5 0

Answer

given,

mass of body = 4 Kg

travelling at speed = 3 m/s

exploded into two fragments of equal mass i.e. = 2 Kg

energy after collision increases by 18 J

original kinetic energy

KE = \dfrac{1}{2}MV^2

KE = \dfrac{1}{2}\times 4 \times 3^2

KE = 18 J

energy increases after collision = 18 J

total energy = 36 J

now,

\dfrac{1}{2}m_1v_1^2 + \dfrac{1}{2}m_2v_2^2 = 36

\dfrac{1}{2}v_1^2 + \dfrac{1}{2}v_2^2 = 36

v_1^2+v_2^2 = 72

using conservation of momentum

m₁v₁ + m₂v₂ = MV

2 v₁ + 2 v₂ = 4 x 3

v₁ + v₂ = 6-------------(1)

squaring both side in the above equation

v₁² + v₂² + 2v₁v₂ = 36

72 + 2v₁v₂ = 36

v₁v₂ = -18-------(2)

putting value of equation (1) in (2)

v₁ (-18 - v₁) = 6

-18 v₁ - v₁² = 6

v₁² + 18 v₁ + 6 = 0

on solving the above equation we get

v₁ = -0.34 m/s

putting value of v₁ in equation (1)

v₂ = 6 + 0.34

v₂ = 6.34 m/s

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Ummm I’m not sure let me do the work
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A train car has a mass of 10,000 kg and is moving at +3.0 m/s. It strikes an identical train car that is at rest. The train cars
shtirl [24]
In order to compute the final velocity of the trains, we may apply the principle of conservation of momentum which is:
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m₁v₁ = m₂v₂

The final mass of the trains will be:
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3 0
3 years ago
A person slaps her leg with her hand, which results in her hand coming to rest in a time interval of 2.75 ms2.75 ms from an init
kumpel [21]

Answer:

2677.3 N

Explanation:

v₀ = initial speed of the hand = 4.75 m/s

v  = final speed of the hand = 0 m/s

m = Total mass of hand and forearm = 1.55 kg

t = time interval for hand to come to rest = 2.75 ms = 0.00275 s

F = Force applied on the leg

Using Impulse-change in momentum equation

F t = m (v - v₀)

F (0.00275) = (1.55) (0 - 4.75)

F = - 2677.3 N

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6 0
3 years ago
Determine the potential difference between the ends of the wire of resistance 5 Ω if 720 C passes through it per minute.
Strike441 [17]

Answer:

The potential difference between the ends of a wire is 60 volts.

Explanation:

It is given that,

Resistance, R = 5 ohms

Charge, q = 720 C

Time, t = 1 min = 60 s

We know that the charge flowing per unit charge is called current in the circuit. It is given by :

I = 12 A

Let V is the potential difference between the ends of a wire. It can be calculated using Ohm's law as :

V = IR

V = 60 Volts

So, the potential difference between the ends of a wire is 60 volts. Hence, this is the required solution.

8 0
3 years ago
If the electron has half the speed needed to reach the negative plate, it will turn around and go towards the positive plate. Wh
In-s [12.5K]

Answer:

 v = -v₀ / 2

Explanation:

For this exercise let's use kinematics relations.

Let's use the initial conditions to find the acceleration of the electron

            v² = v₀² - 2a y

when the initial velocity is vo it reaches just the negative plate so v = 0

           a = v₀² / 2y

now they tell us that the initial velocity is half

          v’² = v₀’² - 2 a y’

          v₀ ’= v₀ / 2

at the point where turn v = 0              

          0 = v₀² /4  - 2 a y '

          v₀² /4 = 2 (v₀² / 2y)  y’

          y = 4 y'

          y ’= y / 4

We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.

         v² = v₀² -2a y’

         v² = 0 - 2 (v₀² / 2y) y / 4

         v² = -v₀² / 4

         v = -v₀ / 2

We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.

7 0
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