A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end. The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it. The mass of the bullet is one-fourth the mass of the rod.(a) What is the final angular speed of the rod? (b) What is the ratio of the kinetic energy of the system after the collision to the kinetic energy of the bullet before the collision?

A) ω = 6v/19L

B) K2/K1 = 3/19

Explanation:

Mr = Mass of rod

Mb = Mass of bullet = Mr/4

Ir = (1/3)(Mr)L²

Ib = MbRb²

Radius of rotation of bullet Rb = L/2

A) From conservation of angular momentum,

L1 = L2

(Mb)v(L/2) = (Ir+ Ib)ω2

Where Ir is moment of inertia of rod while Ib is moment of inertia of bullet.

(Mr/4)(vL/2) = [(1/3)(Mr)L² + (Mr/4)(L/2)²]ω2

(MrvL/8) = [((Mr)L²/3) + (MrL²/16)]ω2

Divide each term by Mr;

vL/8 = (L²/3 + L²/16)ω2

vL/8 = (19L²/48)ω2

Divide both sides by L to obtain;

v/8 = (19L/48)ω2

Thus;

ω2 = 48v/(19x8L) = 6v/19L

B) K1 = K1b + K1r

K1 = (1/2)(Mb)v² + Ir(w1²)

= (1/2)(Mr/4)v² + (1/3)(Mr)L²(0²)

= (1/8)(Mr)v²

K2 = (1/2)(Isys)(ω2²)

I(sys) is (Ir+ Ib). This gives us;

Isys = (19L²Mr/48)

K2 =(1/2)(19L²Mr/48)(6v/19L)²

= (1/2)(36v²Mr/(48x19)) = 3v²Mr/152

Thus, the ratio, K2/K1 =

[3v²Mr/152] / (1/8)(Mr)v² = 24/152 = 3/19

uranium fissions/breaks up giving out energy and neutrons

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards