All categories

2 years ago

Ship 1 has 4 times the mass of ship 2. They both have equal kinetic energy. How much faster( by what factor) is ship 2 moving?

Physics

1 answer:

Stolb23 [73]2 years ago

4 0

Explanation:

Let us assume the mass of ship 2 = K

Mass of ship 1 = 4K

Condition:

Both ships have the same mass

The kinetic energy of a body is the energy due to the motion of it.

It is expressed as:

K.E = $\frac{1}{2} m v^{2}$

m is the mass of the ship

v is the velocity;

Since both ships have equal velocity;

K.E of ship 1 = K.E of ship 2

\frac{1}{2} m v^{2} of ship 1 = \frac{1}{2} m v^{2} of ship 2

multiply through by 2;

$m_{1} v_{1} ^{2} = m_{2} v_{2} ^{2}$

4k x $v_{1} ^{2}$ = k x $v_{2}^{2}$

dividing by k;

4 v_{1} ^{2} = v_{2}^{2}

taking square root of both sides

V₂ = 2V₁

learn more:

Kinetic energy brainly.com/question/6536722

#learnwithBrainly

You might be interested in

1.An elevator is ascending with constant speed of 10 m/s. A boy in the elevator throws a ball upward at 20 m/ a from a height of

laiz [17]

(a) The maximum height reached by the ball from the ground level is 75.87m

(b) The time taken for the ball to return to the elevator floor is 2.21 s

<u>The given parameters include:</u>

constant velocity of the elevator, u₁ = 10 m/s

initial velocity of the ball, u₂ = 20 m/s

height of the boy above the elevator floor, h₁ = 2 m

height of the elevator above the ground, h₂ = 28 m

To calculate:

(a) the maximum height of the projectile

total initial velocity of the projectile = 10 m/s + 20 m/s = 30 m/s (since the elevator is ascending at a constant speed)

at maximum height the final velocity of the projectile (ball), v = 0

Apply the following kinematic equation to determine the maximum height of the projectile.

$v^2 = u^2 + 2(-g)h_3\\\\where;\\\\g \ is \ the \ acceleration \ due \ to\ gravity = 9.81 \ m/s^2\\\\h_3 \ is \ maximum \ height \ reached \ by \ the \ ball \ from \ the \ point \ of \ projection\\\\0 = u^2 -2gh_3\\\\2gh_3 = u^2 \\\\h_3 = \frac{u^2}{2g} \\\\h_3 = \frac{(30)^2}{2\times 9.81} \\\\h_3 = 45.87 \ m$

The maximum height reached by the ball from the ground level (h) = height of the elevator from the ground level + height of he boy above the elevator + maximum height reached by elevator from the point of projection

h = h₁ + h₂ + h₃

h = 28 m + 2 m + 45.87 m

h = 75.87 m

(b) The time taken for the ball to return to the elevator floor

Final height of the ball above the elevator floor = 2 m + 45.87 m = 47.87 m

Apply the following kinematic equation to determine the time to return to the elevator floor.

$h = vt + \frac{1}{2} gt^2\\\\where;\\\\v \ is \ the \ initial \ velocity \ of \ the \ ball \ at \ the \ maximum \ height = 0\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g}} \\\\t = \sqrt{\frac{2\times 47.87}{9.81}} \\\\t = 2.21 \ s$