A bag of sand of mass M is suspended by a string. A bullet of mass m is fired at it with velocity v and gets embedded into it. The loss of kinetic energy in this process is
Text Solution
Verified by ExpertsD
Initial kinetic energy of bullet = $\frac{1}{2}mv^{2}$
After inelastic collision system moves with velocity V
By the conservation of momentum
$mV + 0 = (m + M)V$ ⇒ ⇒ $V = \frac{m \mathbf{v}}{m + M}$
Kinetic energy of system = $\frac{1}{2} (m + M) V^{2}$
= $\frac{1}{2} (m + M) \left(\frac{mV}{m + M}\right)^2$
Loss of kinetic energy = $\frac{1}{2}mv^{2} - \frac{1}{2}(m+M)\left(\frac{mv}{m+M}\right)^{2}$
= $\frac{1}{2} m v^{2} \left( \frac{M}{m+M} \right)$
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems