A pellet of mass m and specific heat capacity | vedclass

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(B) The initial kinetic energy of the pellet is $K = \frac{1}{2} m v^2$.According to the problem,$50\%$ of this kinetic energy is converted into therm

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$\frac{v^2}{4 c}$

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(B) The initial kinetic energy of the pellet is $K = \frac{1}{2} m v^2$.According to the problem,$50\%$ of this kinetic energy is converted into thermal energy $(Q)$ within the pellet.Thus,$Q = 0.5 	imes K = 0.5 	imes \frac{1}{2} m v^2 = \frac{1}{4} m v^2$.The thermal energy gained by the pellet is also given by the formula $Q = m c \Delta T$,where $\Delta T$ is the rise in temperature.Equating the two expressions for $Q$: $m c \Delta T = \frac{1}{4} m v^2$.Solving for $\Delta T$: $\Delta T = \frac{m v^2}{4 m c} = \frac{v^2}{4 c}$.

$A$ pellet of mass $m$ and specific heat capacity $c$ fired from an air gun hits a steel plate with speed $v$. During the impact,$50\%$ of the pellet's kinetic energy is converted to thermal energy in it. The rise in the temperature of the pellet is

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