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Matter Waves

An impulse ...

Question

An impulse $I$ given to a body changes its velocity from $v_{1}$ to $v_{2}$ . The increase in the kinetic energy of the body is given by:

I (v_{1} + v_{2})

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I (v_{1} + v_{2}) / 2

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I (v_{1} - v_{2})

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I (v_{1} - v_{2}) / 2

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Solution

The correct option is A $I (v_{1} + v_{2}) / 2$
Impulse, $I = m (v_{2} - v_{1})$
Change in $K E = \frac{1}{2} m ({v_{2}}^{2} - {v_{1}}^{2}) = \frac{1}{2} m (v_{2} - v_{1}) (v_{2} + v_{1}) = \frac{I (v_{2} + v_{1})}{2}$

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An impulse I given to a body changes its velocity from v1 to v2. The increase in the kinetic energy of the body is given by:

The correct option is A I(v1+v2)/2Impulse, I=m(v2−v1)Change in KE=12m(v22−v12)=12m(v2−v1)(v2+v1)=I(v2+v1)2

An impulse $I$ given to a body changes its velocity from $v_{1}$ to $v_{2}$ . The increase in the kinetic energy of the body is given by:

The correct option is A $I (v_{1} + v_{2}) / 2$
Impulse, $I = m (v_{2} - v_{1})$
Change in $K E = \frac{1}{2} m ({v_{2}}^{2} - {v_{1}}^{2}) = \frac{1}{2} m (v_{2} - v_{1}) (v_{2} + v_{1}) = \frac{I (v_{2} + v_{1})}{2}$