A stone with a mass of 0.100kg rests on a frictionless, horizontal surface. A bullet of mass 2.50g traveling horizontally at 500m/s strikes the stone and rebounds horizontally at right angles to its original direction with a speed of 300m/s. A) Compute the magnitude of the velocity of the stone after it is struck.
B) Compute the direction of the velocity of the stone after it is struck. (degrees from the initial direction of the bullet)

Explanation:

Given that:

mass of stone (M) = 0.100 kg

mass of bullet (m) = 2.50 g = 2.5  ×10 ⁻³ kg

initial velocity of stone () = 0 m/s

Initial velocity of bullet () = (500 m/s)i

Speed of the bullet after collision () = (300 m/s) j

Suppose we represent to be the velocity of the stone after the truck, then:

From linear momentum, the law of conservation can be applied which is expressed as:

∴

The magnitude now is:

Using the tangent of an angle to determine the direction of the velocity after the struck;

Let θ represent the direction:

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.