# A block of mass m = 2.00 kg slides down a 30.0 degree incline which is 3.60 m high. At the...

## Question:

A block of mass m = 2.00 kg slides down a {eq}30.0 ^\circ {/eq} incline which is 3.60 m high. At the bottom, it strikes a block of mass M = 7.20 kg which is at rest on a horizontal surface Determine the speeds of the two blocks after the collision.

## The collision of two objects:

The striking of any two objects in which energy transfer takes place is known as the collision of two objects. Bothe the objects may be in motion, or one object may be rest condition for the collision to take place.

Given data

• The value of the mass of the small block is {eq}m = 2\;{\rm{kg}} {/eq}
• The value of the angle of inclination is {eq}\theta = 30^\circ {/eq}
• The value of the height of the small block is {eq}{h_{small}} = 3.60\;{\rm{m}} {/eq}
• The value of the mass of the big block is {eq}M = 7.20\;{\rm{kg}} {/eq}

The expression for the conservation of energy on the small block to determine the speed of the block at the bottom,

{eq}\begin{align*} P.E &= K.E\\ mgh\sin \theta &= \dfrac{1}{2}m{u_{small}}^2 \end{align*} {/eq}

Substitute the value in the above equation.

{eq}\begin{align*} \left( {9.81} \right)\left( {3.60} \right)\sin {30} &= \dfrac{1}{2}{u_{small}}^2\\ {u_{small}} &= 5.942\;{\rm{m/s}} \end{align*} {/eq}

The expression for the conservation of momentum on striking of the block is,

{eq}m{u_{small}} + M{u_{big}} = m{v_{small}} + M{v_{big}} {/eq}

Substitute the value in the above equation.

{eq}2\left( {5.942} \right) + 7.20\left( 0 \right) = \left( 2 \right){v_{small}} + 7.20{v_{big}}.....................(I) {/eq}

The expression for the conservation of kinetic energy of the blocks is,

{eq}\dfrac{1}{2}m{u_{small}}^2 = \dfrac{1}{2}m{v_{small}}^2 + \dfrac{1}{2}M{v_{big}}^2 {/eq}

Substitute the value in the above equation.

{eq}2{\left( {5.942} \right)^2} = 2{v_{small}}^2 + \left( {7.20} \right){v_{big}}^2.......................(II) {/eq}

Solve the above two equation (I) and (II).

{eq}\begin{align*} {v_{small}} &= - 4.66\;{\rm{m/s}}\\ {v_{big}} &= 3.73\;{\rm{m/s}} \end{align*} {/eq}

Thus, the value of the final velocity of the small block after collision is {eq}- 4.66\;{\rm{m/s}} {/eq} and big block is {eq}3.73\;{\rm{m/s}} {/eq} 