Question

article of mass m has half the K.E of another particle of mass m/2. if the speed of the heavier par
increased by 2 m/s, its new K.E equals the original K.E of the lighter particle then
​

Answer 1

Explanation:

two velocity are possible

v=4/3,v=4

Answer 2

$v_1:v_2=1:2$

Explanation:

Given:

• mass of first body, $m_1=m$

• mass of second body, $m_2=\frac{m}{2}$

• kinetic energy of first body, $KE_1=\frac{1}{2} \times KE$

• kinetic energy of second body, $KE_2=KE$

Also given that:

$KE_1=\frac{1}{2} KE_2$

$\frac{1}{2}\times m\times v_1^2=\frac{1}{2} \times\frac{1}{2} \times \frac{m}{2} \times v_2^2$

$v_1^2=\frac{v_2^2}{4}$

$\frac{v_1}{v_2}=\frac{1}{2}$

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TOPIC: kinetic energy

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