A bullet of mass mB = 0.01 kg is moving with a speed of 120 m/s when it collides
ID: 1906022 • Letter: A
Question
A bullet of mass mB = 0.01 kg is moving with a speed of 120 m/s when it collides with a rod of mass mR = 6 kg and length L = 5 m (shown in the figure). The rod is initially at rest, in a vertical position, and pivots about an axis going through its center of mass. The bullet imbeds itself in the rod at a distance L/4 from the pivot point. As a result, the bullet-rod system starts rotating. (a) Find the angular velocity, ?, of the bulletrod system after the collision. You can neglect the width of the rod and can treat the bullet as a point mass. (b) How much kinetic energy is lost in the collision?
m rnl. mil. LExplanation / Answer
A bullet of mass mB = 0.0127 kg is moving with a speed of 101 m/s when it collides with a rod of mass mR = 7.23 kg and length L = 1.09 m (shown in the figure). The rod is initially at rest, in a vertical position, and pivots about an axis going through its center of mass. The bullet imbeds itself in the rod at a distance L/4 from the pivot point. As a result, the bullet-rod system starts rotating. a) Find the angular velocity, ?, of the bullet-rod system after the collision. You can neglect the width of the rod and can treat the bullet as a point mass. b) How much kinetic energy is lost in the collision? (Hint: Please enter a positive number for your answer! If ?K < 0, then the amount of kinetic energy lost is a positive number.) http://www.hizliupload.com/viewer.php?file=94332335472721891326.png In this case the rod will not translate, it will only rotate. So, to determine ?, L (Angular Momentum) must be conserved about the hinge, since no external torque acts on that point. Initially, L B (angular momentum of bullet about the hinge) = mB*v*perpendicular distance = (mB)vL/4. Finally, L S (L system) = I(system)?. [ I = moment of inertia] I system = mR * L² /12 + mB * (L/4)² [assuming the rod to be uniform, its COM is at its centre. I about that point = m l² /12) = mR * L² /12 + mB * L² / 16 = L² [mR/12 + mB/16]. Conserving L, (mB) v L / 4 = L² [mR/12 + mB/16]? => mB v = L (mR / 3 + mB /4) ? => ? = mB v / L (mR / 3 + mB /4) . Putting values, ? = 0.0127*101/1.09(7.23/3 + 0.0127/4) rad/s = 1.2827/2.63 = 0.488 rad/s (Answer). K Initial = 1/2 mB v² = 0.5*0.0127*)(101)² = 64.77635 Joule. K Final = 1/2 I ?² = 0.5 * 0.6 * (1.09)² * (0.488)² = 0.17 Joule. So, Energy lost = 64.6 Joule (Answer)Related Questions
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