Question

Block A in the figure (Figure 1) has mass 1.00 kg, and block has mass 3.00 kg . The blocks are forced together, compressing a spring S between them; then the system is released from rest on a level, frictionless surface. The spring, which has negligible mass, is not fastened to either block and drops to the surface after it has expanded. Block B acquires a speed of 1.20m/s .


A) What is the final speed of block ?
- I tried 0.4 m/s, which was incorrect

B)How much potential energy was stored in the compressed spring?

Answer 1

Concepts and reason

The concept required to solve the given problem is the law of conservation of linear momentum and law of conservation of energy.

Calculate the final speed of block A by the law of conservation of momentum.

Calculate the potential energy stored in the spring by the law of conservation of energy.

Fundamentals

Law of conservation of linear momentum: It states that when no external force acts on a body the momentum remains conserved. Mathematically the statement may be expressed as,

${p_i} = {p_f}$

Here, ${p_i}{\rm{ and }}{p_f}$ is the initial and final momentum of the body into consideration.

Momentum is given by,

$p = mv$

Here, $m$ is the mass and $v$ is the velocity of the body.

Law of conservation of energy: It states that the energy can neither be created nor destroyed but can only be changed from one form to another. Mathematically it may be written as,

$U + K = {\rm{ constant}}$

Or,

$\begin{array}{c}\\{U_f} + {K_f} = {U_i} + {K_i}\\\\{U_i} - {U_f} = {K_f} - {K_i}\\\\\left| {\Delta U} \right| = \left| {\Delta K} \right|\\\end{array}$

Here, ${U_f}{\rm{ and }}{U_i}$ are the final and initial potential energy respectively and ${K_f}{\rm{ and }}{K_i}$ are the initial and final kinetic energy respectively.

(A)

Apply the conservation of linear momentum.

${m_A}{u_A} + {m_B}{u_B} = {m_A}{v_A} + {m_B}{v_B}$

Here, ${m_A}{\rm{ and }}{m_B}$ are the masses of block A and B, ${u_A}{\rm{ and }}{u_B}$ are the initial velocities of blocks A and B and ${v_A}{\rm{ and }}{v_B}$ are the final velocities of block A and B.

Since both the blocks are at rest initially hence the initial speed of both the blocks will be $0{\rm{ m / s}}$ .

Substitute $1.00{\rm{ kg}}$ for ${m_A}$ , $3.00{\rm{ kg}}$ for block B, $0{\rm{ m / s}}$ for ${u_A}{\rm{ and }}{{\rm{u}}_B}$ and $1.20{\rm{ m / s}}$ for ${v_B}$ in the above equation.

$\begin{array}{c}\\\left( {1.00{\rm{ kg}}} \right)\left( {0{\rm{ m / s}}} \right) + \left( {3.00{\rm{ kg}}} \right)\left( {0{\rm{ m / s}}} \right) = \left( {1.00{\rm{ kg}}} \right){v_A} + \left( {3.00{\rm{ kg}}} \right)\left( {1.20{\rm{ m / s}}} \right)\\\\{v_A} = \frac{{\left( {3.00{\rm{ kg}}} \right)\left( {1.20{\rm{ m / s}}} \right)}}{{\left( {1.00{\rm{ kg}}} \right)}}\\\\ = 3.6{\rm{ m / s}}\\\end{array}$

The final velocity of block A will be $3.6{\rm{ m / s}}$ .

(B)

The potential energy stored in the spring will be equal to the kinetic energies of the two blocks according to the law of conservation of energy.

Apply the law of conservation of energy.

${U_{{\rm{Spring}}}} = {K_A} + {K_B}$

Here, ${U_{{\rm{Spring}}}}$ is the potential energy stored in the spring, ${K_A}$ is the kinetic energy of block A and ${K_B}$ is the kinetic energy of block B.

Substitute $\frac{1}{2}{m_A}{v_A}^2$ for ${K_A}$ and $\frac{1}{2}{m_B}{v_B}^2$ for ${K_B}$ in the above equation.

${U_{{\rm{Spring}}}} = \frac{1}{2}{m_A}{v_A}^2 + \frac{1}{2}{m_B}{v_B}^2$

Substitute $1.00{\rm{ kg}}$ for ${m_A}$ , $3.00{\rm{ kg}}$ for block B, ${\rm{3}}{\rm{.6 m / s}}$ for ${v_A}$ and $1.20{\rm{ m / s}}$ for ${v_B}$ in the above equation.

$\begin{array}{c}\\{U_{{\rm{Spring}}}} = \frac{1}{2}\left( {1.00{\rm{ kg}}} \right){\left( {{\rm{3}}{\rm{.6 m / s}}} \right)^2} + \frac{1}{2}\left( {3.00{\rm{ kg}}} \right){\left( {{\rm{1}}{\rm{.2 m / s}}} \right)^2}\\\\ = 8.64{\rm{ J}}\\\end{array}$

Ans: Part A

The final speed of block A will be $3.6{\rm{ m / s}}$ .