Applying the law of conservation of momentum

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• Physics

  = 11.85 Kg.m/s B) The second problem can be solved by applying the Law of Conservation of Energy to the system of two moving carts.

• Collision

  Law of conservation of linear momentum : When two bodies collide, their momentum is conserved. That is, total momentum of the bodies before and after collision is same.

• Physics: Muzzle velocity of a rifle. What is the speed of the bullet.

  The oscillation period is 1.25 seconds (just in case it's hard to read) Applying law of conservation of momentum, mv+M*0 = (m+M) v2 ------------------1 where m = mass of bullet = 7.87 g M = mass of block = 6.085 Kg v2 = m*v/(m+M) Applying

• Inelastic collisions

  Law of conservation of momentum is applied in the solution.

• Angular Momentum

  after the rings leave the rod = ω2 Again applying conservation of angular momentum : [(1/12)m1L2 + 2m2r2] x ω = (1/12)m1L2 ω2 ω2 = ω x[(1/12)m1L2 + 2m2r2]/[(1/12)m1L2] This problem is a good example to understand the principle of conservation

• Newton's laws of motion: Collision

  There is no tangential external force on the system of the two objects and hence according to the law of conservation of momentum the momentum of the system will remain unchanged.

• Mechanics: Motion of a ball rolled on incline and projectile

  we have Final momentum of the system = Initial momentum Or mv1 + mv2 = m*1.4 + m*0 Or v1 + v2 = 1.4 ---------- (1) Now applying the law of conservation of energy we have ½ mv12 + ½ mv22 = ½ m*1.42 or v12 + v22 = 1.42 ------

• Newton's 3rd Law

  f) smaller in magnitude and parallel to the force on the earth due to the moon. Newton's 3rd Law is a consequence of conservation of momentum.

• Spinning satellites

  1 rpm = [2(pi) / 60] rad/s ) For the Space craft, Moment of inertia (Is) = 500 Kg-m^2 a) to calculate the spin rate, we need to apply the Law of conservation of angular momentum between the space craft and the wheel.