Question

Two particles of equal mass m at a and b are connected rigid light rod ab lying table. An impulse j is applied at a in the plane of the table and perpendicular at ab. Then the velocity of the particle at a i

Answer 1

Answer:

Speed of the particle at position A is given as

$v_A = \frac{J}{m}$

Explanation:

As we know that impulse is applied at particle A

now by impulse momentum equation we can say that total momentum transferred to the center of mass is equal to the impulse given at A

$J = (2m)v$

$v = \frac{J}{2m}$

Now by angular momentum equation we have

$J(\frac{L}{2}) = I\omega$

here we have

$I = \frac{mL^2}{4} + \frac{mL^2}{4}$

$I = \frac{mL^2}{2}$

so we have

$J(\frac{L}{2}) = \frac{mL^2}{2}\omega$

$\omega = \frac{J}{mL}$

now speed of particle A is given as

$v_A = v + \frac{\omega L}{2}$

$v_A = \frac{J}{2m} + \frac{J}{2m}$

$v_A = \frac{J}{m}$

#Learn

Topic : Impulse momentum theorem

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