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Conservation of Momentum: 1 Dimension Collision : AskPhysics

'Two objects of masses m and 3m undergo a collision in one dimension. ... velocity before the collision: [; m3v - 3mv = m * v_1 + 3m*v_2 ;] Then solving for the ... So, the momentum of the two objects before they collide is equal in magnitude but ... I have been experiencing a lot of anxiety with my math background since I took ...

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Pseudovector - Wikipedia

... electric field is) are "true" vectors, but the magnetic field B is a pseudovector. In physics and mathematics, a pseudovector (or axial vector) is a quantity that transforms like a ... The vector a × b is a normal to the plane (there are two normals, one on each side – the ... Consider the pseudovector angular momentum L = r × p.

For more information, see Pseudovector - Wikipedia


Given that

  Object A velocity = v_1

  Object B velocity = v_2

  Momentum of object A (p_1) = (p_2) Momentum of object B

  Inequality in velocities is   v_1 < v_2 ...........................(1)

Step 1: Recall the formula of momentum and kinetic energy

              Momentum p = mv

              Kinetic energy K.E = \frac{1}{2} mv^2

                   Where, p - momentum, m - mass and v - velocity  

              Momentum of object A,   p_1 = m_1 v_1

              Momentum of object B, p_2 = m_2 v_2   


              K.E of object A,   K.E_A = \frac{1}{2} m_1 (v_1)^2

              K.E of object B,   K.E_B = \frac{1}{2} m_2 (v_2)^2

Step 2: Multiplying the equal values(constants) on both sides of the equation (1)

                       v_1 < v_2

                       \frac{1}{2} v_1 < v_2 \frac{1}{2}                           

                       \frac{1}{2} p_1 * v_1 < p_2 * v_2 \frac{1}{2}                              \because p_1 = p_2

                       \frac{1}{2} p_1 * v_1 < p_2 * v_2 \frac{1}{2}


Step 3: Substitute the values of p_1 \text{ and } p_2 . in the above equation.

                [math] \frac{1}{2} [m_1 v_1] * v_1 < [m_2 v_2] * v_2 \frac{1}{2} [/math]

                [math] \frac{1}{2} [m_1 (v_1)^2 < m_2 (v_2)^2 * \frac{1}{2} [/math]

                   K.E_A < K.E_B

            Hence, kinetic energy of B is greater than kinetic energy A

                          Option D is the correct answer.