PHYSICS PROBLEM - Assignment Example

Physics Problem A bullet of mass m = 8.00g is fired against a block of Mass M that is initially at rest at the edge of a table of height=1.00m. The bullet remains in the block and after the impact, the block lands d=1.90m from the bottom of the table. To determine the initial speed of the bullet, one must understand the following. This is a form of an inelastic collision where the bullet comes to rest inside the block. Therefore, kinetic energy and linear momentum are lost.Initially, the block being at rest has zero kinetic energy but has a potential energy= mgh, i.e., m=mass, g=gravity and h=height.

On collision, the bullet and the block become one and keep moving together for some time t until they fall to the floor in a projectile like motion. M1V1 + M2V2 = M3 (V1 + V2) M1= mass of bullet, M2= mass of block M3 = Total mass of block and bullet2. The physical quantities and variables here are mass, momentum, velocity, height and projectile motion.3. The equations to be used in this problem are: i) M1V1 + M2V2 = M3 (V1 + V2) i.e., the from the law of conservation of linear momentumii) S = ut + 1/2at2 and V= U2 + 2as The 2nd and the 3rd Equation of linear motion 4.

SolutionCommon velocity From the 2nd equation of linear motion, S= ut + 1/2at2 and = 0 1.90= 1/2(10)t2 1.90 = 5t2 t = 0.6164s velocity = d/t = 1.90/0.6164 = 3.0822m/s substituting this in equation (I) 0.008 (v1) = 0.248(3.0822) v1 = 95.548m/sWork CitedBoldea, I, and S.A Nasar. Linear Motion Electromagnetic Systems. New York: Wiley-Interscience, 1985. Print.