WebFeb 25, 2016 · Both the bullet and the block are initially at 25.0°C. Assume that no heat is lost to the surroundings and that all the kinetic energy; As shown in the diagram, a bullet of mass 7.0 g strikes a block of wood of mass 2.1 kg. The block of wood is suspended by a string of length 2.0 m, forming a pendulum. WebA 15g bullet strikes and becomes embedded in a 1.25kg block of wood placed on a horizontal surface just in front of the gun. If the coefficient of kinetic friction (µ) between …
g A 24-gg bullet strikes and becomes embedded in a 1.50-kgkg block …
WebApr 25, 2024 · The original velocity of the bullet is 909 m/s.. Let be the mass of bullet and be the mass of wooden block.. And also let V be the final velocity of the bullet and block and be the velocity of the bullet.. Given, will be 0.05 kg and will be 5 kg. Final velocity V is 9 m/s.. we have to calculate initial velocity of the bullet.. From the principle of conservation … WebDec 9, 2024 · The muzzle speed of the bullet before the collision is 415.3 m/s. The given parameters: Mass of the bullet, m₁ = 24 g; Mass of the wood, m₂ = 1.5 kg; Coefficient of kinetic friction, μk = 0.23; Distance traveled by the block before stopping, d = 9.5 m; Apply the principle of work-energy theorem to determine the final velocity of the block ... dr nazri usm
Worked example 6.4: Bullet and block - University of Texas at Austin
WebGet an answer for 'A 15g bullet strikes and becomes embedded in a 1.25kg block of wood placed on a horizontal surface just in front of the gun. If the coefficient of kinetic friction (µ) between ... http://people.tamu.edu/~mahapatra/teaching/ch8_mp_sols.pdf WebLet the velocity of the embedded block be V. Given: the mass of bullet is m, and mass of the wooden block is M, initial velocity of the bullet be v. Using law conservation of momentum: m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2 Initially the block is at rest. So u 1 = 0 m s − 1 ∴ Total momentum before impact = m v Finally the bullet is embedded ... raouf rzaev