Question

S GP A projectile of mass m moves to the right with a speed vi (Fig. P11.51a). The projectile strikes and sticks to the end of a stationary rod of mass M , length d , pivoted about a frictionless axle perpendicular to the page through O (Fig. P11.51b). We wish to find the fractional change of kinetic energy in the system due to the collision.(g) What is the kinetic energy of the system after the collision?

Answer 1

What is the kinetic energy of the system after the collision?

$K_f=\frac{3}{2} \frac{m^{2}v_i^{2} }{(M+3m)}$

How this is calculated?

Given:

Initial speed=$v_i$

mass of rod=M

Let, Initial kinetic energy =$K_i$

Final kinetic energy=$K_f$

Moment of inertia =I

What is the moment of inertia?

$I=(I_p)_0+(I_{rod})_0\\I=m(\frac{d}{2})^{2} +\frac{Md^{2} }{12} \\I=\frac{(M+3m)d^{2} }{12}$

What is the angular momentum?

By conservation of angular momentum,

$L_i=L_f$

$mv_i\frac{d}{2}=\frac{(M+3m)d^{2}\omega }{12} \\\omega=\frac{6mv_i^{2} }{d(M+3m)}$

We know that, the final kinetic energy is given by,

$K_f=I\omega^{2}\\K_f=\frac{1}{2} *\frac{(M+3m)d^{2} }{12} *\frac{36m^{2}v_i^{2}}{d^{2}(M+3m)^{2}}\\ K_f=\frac{3}{2} \frac{m^{2}v_i^{2} }{(M+3m)}$

What is the kinetic energy?

• In physics, the kinetic energy of an object is the energy that it possesses due to its motion.
• It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.
• Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes.

To know more about kinetic energy, refer:

brainly.com/question/114210

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