Question

The track shown in the figure is frictionless. The block B of mass 4 kg is lying at rest and block A of mass 2 kg is pushed along the track with some speed. The collision between A and B is perfectly elastic.With what velocity should the block A start such that block B just reaches at point P?

• (A) 10 m/s
• (B) 5 m/s
• (C) 20 m/s
• (D) 15 m/s

Answer 1

The Correct Option is A

Explanation:
Given:Mass of the block A, mA = 2 kgMass of the block B, mB = 4 kgHeight of block A from the ground, = 4 mThe collision between block A and block B is perfectly elastic.We have to find the velocity with which the block A should start such that block B just reaches the point P.Let the block A starts with velocity uA and it reaches to block B with the velocity vA.Applying the law of conservation of energy for block A, we have$\frac{1}{2}{m}_A{u}_A^2+m_{A}gh=\frac{1}{2}{m}_A{v}_A^2$$\Rightarrow \frac{{\mu }^{2}A}{2}+10\times 4=\frac{v^{2}A}{2}$$\Rightarrow \frac{{\mu }_A^2+80}{2}=\frac{v_A^2}{2}$$\Rightarrow v_A^2={\mu }_A^3+80$....(i)This is the velocity with which block A collides with the block B.Applying the law of conservation of linear momentum just before and after the collision, we have$m_A{v}_A+0=m_A{v}_A+m_B{u}_B,$[Initial momentum of the block B is zero as it is at rest.]where,VA : final velocity of the block AUB : velocity of the block B after the collision$\Rightarrow 2v_A=2V_A=4u_B$$\Rightarrow v_A=2u_B+V_A$.......(ii)Now, as the collision between the blocks A and B is elastic, applying the law of conservation of kinectic energy just before and after the collision. We have,$\frac{1}{2}{m}_A{v}_A^2=\frac{1}{2}{m}_A{v}_A^2+\frac{1}{2}{m}_B{u}_B^2$$\Rightarrow v_A^2=2u_B^2+V_A^2$.....(iii)Substituting $v_A$ from eq. (iii) in eq. (ii), we get$v_A=2u_B+\sqrt{v_A^2-2u_B^2}$$\Rightarrow v_A-2u_B=\sqrt{v_A^2-2u_B^2}$$\Rightarrow v_A^2+4u_B^2-4v_A{u}_B=v_A^2-2u_B^2$ [squaring both sides]$\Rightarrow 4u_B^2+2u_B^2=4v_A{u}_B$$\Rightarrow 6u_B^2=4v_A{u}_B$$\Rightarrow 6u_B=4v_A$$\Rightarrow u_B=\frac{2}{3}{v}_A$.....(iv)Substituting $v_A$ from eq. (i) in eq. (iv), we get$u_B=\frac{2}{3}\sqrt{u_A^2+80}$......(v)This is the velocity of the block B with which it moves after the collision.To reach just at the point P, all of the kinetic energy of the block B gets converted into its poteintial energy .$\Rightarrow \frac{1}{2}{m}_B{u}_B^2=m_{B}gh$$\Rightarrow u_B^2=2gh$$\Rightarrow \frac{4}{9}(u_A^2+80)=2\times 10\times 4$ .......[using eq .(v)]$\Rightarrow u_A^2+80=\frac{2\times 10\times 4\times 9}{4}$$\Rightarrow u_A^2=100m^2/s^2$$\Rightarrow u_A=10m/s$This is the velocity with which the block A should start.Hence, the correct option is (A).

Answer 2

The interaction between the two bodies due to which the direction and magnitude of the velocity of the colliding bodies changes are called a collision.

• A collision occurs when two bodies come in direct contact with each other.
• It is a situation in which two or more bodies exert forces on each other in about a relatively short time.

Elastic Collision

If in a particular collision, there is no dissipation of energy, the total kinetic energy of the objects before collision is equal to the total kinetic energy of the objects after collision. Such a collision is termed an Elastic collision.

• A collision between two particles is said to be a perfectly elastic collision if both the linear momentum and kinetic energy of the system remain conserved.
• The value of the coefficient of restitution in a perfectly elastic collision is, e = 1

Inelastic Collision

If, in a particular collision, there is a dissipation of energy, the total kinetic energy of the objects before and after collision is not conserved. Such a collision is termed an inelastic collision.

• A collision is said to be a perfectly inelastic collision if two colliding bodies after collision stick together and moves as one body.
• In this collision, the linear momentum of the system remains conserved but the kinetic energy is not conserved.
• The value of the coefficient of restitution in a perfectly inelastic collision is, e = 0.