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grin007 [14]
3 years ago
12

A 15.0-kg object sitting at rest is struck elastically in a head-on collision with a 10.5-kg object initially moving at 3.0 m/s.

Find the final velocity of the 15.0-kg object after the collision.
Physics
1 answer:
DIA [1.3K]3 years ago
7 0

Answer:

The final velocity of the 15.0-kg object after the collision is 2.47 m/s in forward direction.

Explanation:

Given;

mass of the object, m₁ = 15 kg

initial velocity of this object, u₁ = 0

mass of the second object, m₂ = 10.5 kg

initial velocity of this object, u₂ = 3.0 m/s

let the final velocity of the first object = v₁

also, let the final velocity of the second object = v₂

Apply the principle of conservation of linear momentum

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(15 x 0) + (10.5 x 3) = 15v₁ + 10.5v₂

31.5 = 15v₁ + 10.5v₂ ----- (1)

One directional velocity;

u₁ + v₁ = u₂ + v₂

0 + v₁ = 3 + v₂

v₂ = v₁ - 3   ------(2)

Substitute (2) into (1);

31.5 = 15v₁ + 10.5v₂

31.5 = 15v₁  + 10.5(v₁ - 3)

31.5 =  15v₁   + 10.5v₁ - 31.5

63 = 25.5v₁

v₁ = 63 / 25.5

v₁ = 2.47 m/s

Therefore, the final velocity of the 15.0-kg object after the collision is 2.47 m/s in forward direction.

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A car is being tested for safety by colliding it with a brick wall.The 1300 kg car is initially driving towards the wall at a sp
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Answer:

44200 N

Explanation:

To calculate the average force exerted on the car, we will use the following equation

\begin{gathered} Ft=\Delta p \\ Ft=m(v_f-v_i) \end{gathered}

Where F is the average force, t is the time, m is the mass, vf is the final velocity and vi is the initial velocity of the car.

Replacing t = 0.5s, m = 1300 kg, vf = -2 m/s, and vi = 15 m/s and solving for F, we get

\begin{gathered} F(0.5s)=(1300\text{ kg\rparen\lparen-2 m/s - 15 m/s\rparen} \\ F(0.5s)=(1300\text{ kg\rparen\lparen-17 m/s\rparen} \\ F(0.5s)=-22100\text{ kg m/s} \\ F=\frac{-22100\text{ kg m/s}}{0.5\text{ s}} \\ F=-44200\text{ N} \end{gathered}

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Answer:

F = \frac{Qq}{2\pi \epsilon_0 L d}

Explanation:

As we know that if a charge q is distributed uniformly on the line then its linear charge density is given by

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now the electric field due to long line charge at a distance d from it is given as

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now the force on the other charge in this electric field is given as

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