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

Answer:

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

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.

$KE_i + PE_i = KE_f + PE_f\\\frac{1}{2}m_1v_1^2+m_1gh_1 = \frac{1}{2}m_1v_{1f}^2+m_1gh_{1f}$ where $h_{1f} = 0$

So,

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

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

$v_{1f} = \frac{m_1-m_2}{m_1+m_2}*v_1$

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

$v_{1f} = \frac{2.03-4.97}{2.03+4.97} * 5.217 = -2.1911$ $m/s$

(-) 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.

$v_{2f} = \frac{2m_1}{m_1+m_2}*v_1$. Here, $v_1$ will be  $v_{1f}$ found in part (a) again.

$v_{2f} = \frac{2 * 2.03}{2.03+4.97} * 5.217 = 3.0259$ $m/s$

(+) 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.

$\frac{1}{2}mv^2+mgh = \frac{1}{2}mv_f^2+mgh_f$ where $h = 0$ and $v_f = 0$

So,

$gh_f = \frac{1}{2}v_1^2$

$h_f = \frac{v_1^2}{2g} = \frac{(-2.1911)^2}{2*9.8} = 0.2449$ $m$