Question

Starting with an initial speed of 4.54 m/s at a height of 0.337 m, a 2.03-kg ball swings downward and strikes a 4.97-kg ball that is at rest, as the drawing shows.

(a) Using the principle of conservation of mechanical energy, find the speed of the 2.03-kg ball just before impact.
(b) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 2.03-kg ball just after the collision.
(c) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 4.97-kg ball just after the collision.
(d) How high does the 2.03-kg ball swing after the collision, ignoring air resistance

Answer 1

a) 5.2170 m/s

b) 2.1911 m/s, to the left

c) 3.0259 m/s, to the right

d) 0.2449 m

Step-by-step explanation:

Given parameters:

`v_1 = 4.54`
`m/s`

`h_1 = 0.337`
`m`

`m_1 = 2.03`
`kg`

`m_2 = 4.97`
`kg`

`v_2 = 0`
`m/s`

a) Here, the conservation of energy principle can be used.

`KE_i + PE_i = KE_f + PE_f\\(1)/(2)m_1v_1^2+m_1gh_1 = (1)/(2)m_1v_(1f)^2+m_1gh_(1f)` where
`h_(1f) = 0`

So,

`(1)/(2)v_1^2+gh_1 = (1)/(2)v_(1f)^2\\v_(1f) = \sqrt{2((1)/(2)v_1^2+gh_1)} = \sqrt{2((1)/(2)*4.54^2+9.8*0.337)} = 5.2170`
`m/s`

b) Here, the conservation of momentum principle can be used.

`v_(1f) = (m_1-m_2)/(m_1+m_2)*v_1`

Here, our new
`v_1` will be
`v_(1f)` found in part (a).

`v_(1f) = (2.03-4.97)/(2.03+4.97) * 5.217 = -2.1911`
`m/s`

(-) sign means that the 2.03 kg ball after collision will bounce back in the left direction.

c) Here, we can use the conservation of momentum principle again.

`v_(2f) = (2m_1)/(m_1+m_2)*v_1`. Here,
`v_1` will be
`v_(1f)` found in part (a) again.

`v_(2f) = (2 * 2.03)/(2.03+4.97) * 5.217 = 3.0259`
`m/s`

(+) sign means that the 4.97 kg ball after collision will swing to the right direction.

d) Here, we can use the conservation of energy principle again.

`(1)/(2)mv^2+mgh = (1)/(2)mv_f^2+mgh_f` where
`h = 0` and
`v_f = 0`

So,

`gh_f = (1)/(2)v_1^2`

`h_f = (v_1^2)/(2g) = ((-2.1911)^2)/(2*9.8) = 0.2449`
`m`