Question

A completely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axis. Just before the collision, one ball, of mass 3.8 kg, is moving upward at 19 m/s and the other ball, of mass 1.6 kg, is moving downward at 11 m/s. How high do the combined two balls of putty rise above the collision point? (Neglect air drag.)

Answer 1

5.21m

Step-by-step explanation:

We are given that;

mass of first ball; m1 = 3.8kg

Speed of first ball; v1 = 19 m/s upwards

Mass of second ball; v2 = 1.6 kg

Speed of second ball; v2 = 11 m/s downwards

From conservation of linear momentum,

m1v1 + m2v2 = m_t•v_t

Where,

m1v1 is momentum of first ball

m2v2 is momentum of second ball

m_t•v_t is the combined momentum of the 2 balls.

Let's make v_t the subject

v_t = [m1v1 + m2v2]/m_t

m_t is the combined mass of both balls.

Since first and second ball are moving in opposite directions, thus;

v_t = [m1v1 - m2v2]/m_t

Thus, m_t = 3.8 + 1.6 = 5.4kg

Thus, plugging in the relevant values, we have;

v_t = [(3.8 x 19) - (1.6 x 11)]/5.4

v_t = [(3.8 x 19) - (1.6 x 11)]/5.4

v_t = (72.2 - 17.6)/5.4

v_t = 54.6/5.4 = 10.11 m/s

Now, from equation of motion,

v² = u² + 2gh

Where u is initial velocity which is now v_t while final velocity v is zero.

Now, since gravity is acting against motion, g = - 9.8m/s²

Thus,

v² = u² + 2gh gives;

0² = 10.11² - (2 x 9.8 x h)

19.6h = 102.212

h = 102.212/19.6

h = 5.21 m

Answer 2

`\Delta h = 5.212\,m`

Step-by-step explanation:

The inelastic collision is modelled by using the Principle of Momentum Conservation:

`(3.8\,kg)\cdot (19\,(m)/(s) ) + (1.6\,kg)\cdot (-11\,(m)/(s) ) = (3.8\,kg + 1.6\,kg)\cdot v`

The final velocity is:

`v = 10.111\,(m)/(s)`

The maximum height of the composite system is:

`(0\,(m)/(s))^(2) = (10.111\,(m)/(s) )^(2) - 2\cdot (9.807\,(m)/(s^(2)) )\cdot \Delta h`

`\Delta h = 5.212\,m`