Answer to Question #4623 in Mechanics | Relativity for manikarthik
a bullet of mass 0.01kg travelling at a speed of 500meters per seconds strikes a block of mass 2kg which is suspended by a string of length 5 meters and emerges out.The block rises by a vertical heigth of 0.1meters.What is the speed of bullet?
Let the m be the mass of a bullet, M the mass of a block, V velocity of a bullet, Vb velocity of block, L - the length of a string, A - an angle between the vertical axis and the string, h - a height of rising. We'll use the energy conservation law. (mV^2)/2 = (M*Vb^2)/2 (*) Also, Vb^2 = 2*g*L*(1-cosA) = 2*g*L*(1-(L-h)/L) = 2*9.8*5*(1-(5-0.1)/5) = 1.96 (m/s)^2. From (*): V = sqrt((M*Vb^2)/m) = sqrt((2*1.96)/0.01) = 19.79 m/s.