Question

Two objects collide and stick together. What will happen to the total kinetic energy of the objects (increase, decrease, stays the same)? Explain what happens to the kinetic energy.Effect on total kinetic energy: Description of Kinetic Energy changes:

Answer 1

Given:

Two objects collide and stick together.

To find:

Effect on the kinetic energy due to the collision.

Step-by-step explanation:

When perfectly inelastic bodies moving along the same line collide, they stick together. An inelastic collision is a collision in which there is a loss in kinetic energy.

Let m1 and m2 be the masses and velocity v1 and v2 be the velocities of objects along the same line before the collision.

Let V be the velocity of objects after they collide and stick together.

By the law of conservation of linear momentum,

`m_1v_1+m_2v_2=m_1V+m_2V`

Rearranging the above equation, we get:

`V=(m_1v_1+m_2v_2)/(m_1+m_2)\text{ ......\lparen1\rparen}`

The total kinetic energy of objects before the collision is:

`\text{Kinetic Energy \lparen before collision\rparen}=m_1v_1^2+m_2v_2^2\text{ ......\lparen2\rparen}`

The kinetic energy after collision is:

`\text{Kinetic energy \lparen after collision\rparen}=(1)/(2)(m_1+m_2)V^2\text{ ......\lparen3\rparen}`

Using equations (1), (2), and (3), the loss in kinetic energy is given as:

`\begin{gathered} (1)/(2)m_1v_1^2+(1)/(2)m_2v_2^2-(1)/(2)(m_1+m_2)V^2=(1)/(2)\lbrack m_1v_1^2+m_2v_2^2-(m_1v_1+m_2v_2)/(m_1+m_2)\rbrack \\ (1)/(2)m_1v_1^2+(1)/(2)m_2v_2^2-(1)/(2)(m_1+m_2)V^2=(1)/(2)\lbrack(m_1m_2\left(v_1^2+v_2^2-2v_1v_2\right))/(m_1+m_2)] \\ (1)/(2)m_1v_1^2+(1)/(2)m_2v_2^2-(1)/(2)(m_1+m_2)V^2=(m_1m_2\left(v_1-v_2\right)^2)/(2(m_1+m_2)) \end{gathered}`

From the above result, we see that the energy loss in the kinetic energy is positive.

Thus, the kinetic energy of decreases, when two objects collide and stick together.

Final answer:

The kinetic energy of decreases, when two objects collide and stick together.