# A block of mass 4.253 kg is released on the track at a height 5.52 m above the level surface. It...

## Question:

A block of mass 4.253 kg is released on the track at a height 5.52 m above the level surface. It slides down the track and makes a head-on elastic collision with a block of mass 17.012 kg, initially at rest. The acceleration of gravity is 9.8 m/s{eq}^2 {/eq}. Calculate the height to which the block with mass 4.253 kg rises after rebounding from the collision.

## Conservation of Energy:

The energy conservation principle consists of a physical law that states that if no external forces act on the system, the total energy of the system will conserve. Energy is not created nor destroyed, it only transforms. This means that a system will present energy transformations, but the overall total energy of the system will always be the same. The total energy of a system can be determined by the sum of the total kinetic and potential energy of the system.

$$E_t=K_e+U_e\\ E_t=0.5mv^2+mgh $$

## Answer and Explanation:

We use the energy conservation principle to determine the velocity at which the collides with the other mass. The initial potential energy of the mass will be equal to the final kinetic energy of the block.

{eq}E_i=E_f\\ mgh=\dfrac{1}{2}mv^2\\ v=\sqrt{2gh}\\ v=\sqrt{2\times9.8\times 5.52}\\ v= 10.40\ m/s {/eq}

As the block collides elastically with the other block, we use momentum conservation to determine the rebound speed of the block.

{eq}m_1v_1=m_1v_2+m_2v_3\\ v_3=\dfrac{m_1v_1-m_1v_2}{m_2} {/eq}

As the collision is elastic, the total kinetic energy of the system will conserve.

{eq}\dfrac{1}{2}m_1v_1^2=\dfrac{1}{2}m_1v_2^2+\dfrac{1}{2}m_2v_3^2\\ m_1v_1^2=m_1v_2^2+m_2v_3^2\\ m_1v_1^2=m_1v_2^2+m_2(\dfrac{m_1v_1-m_1v_2}{m_2})^2\\ v_1^2=v_2^2+\dfrac{m_1v_1^2-2m_1v_1v_2+m_1v_2^2}{m_2}\\ v_1^2=v_2^2+\dfrac{m_1}{m_2}(v_1^2-2v_1v_2+v_2^2)\\ 10.40^2=v_2^2+\dfrac{4.253}{17.012}(10.40^2-2\times10.40v_2+v_2^2)\\ 108.16=v_2^2+27.04-5.2v_2+0.25v_2^2\\ 1.25v_2^2-5.2v_2-81.12=0\\ \dfrac{5.2\pm \sqrt{5.2^2-4\times1.25\times-81.12}}{2.50}\\ v_2=-6.24\ m/s {/eq}

Finally, we use energy conservation to find the height.

{eq}E_i=E_f\\ \dfrac{1}{2}mv_2^2=mgh_2\\ h_2=\dfrac{v_2^2}{2g}\\ h_2=\dfrac{6.24^2}{2\times9.8}\\ h_2= 1.986\ m {/eq}

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from ICSE Environmental Science: Study Guide & Syllabus

Chapter 1 / Lesson 6