A table tennis ball is dropped unto the floor from a height of 4m and it rebounds to a height of 3m. If the time of contact with the floor is 0.01secs, what is the magnitude and direction of the acceleration during the contact
To find the magnitude and direction of the acceleration during the contact, we can use the basic kinematic equation:
\[v_f = v_i + at\]
Where:
- \(v_f\) is the final velocity (in this case, the ball's velocity just before it rebounds)
- \(v_i\) is the initial velocity (in this case, the ball's velocity just before it hits the floor)
- \(a\) is the acceleration
- \(t\) is the time of contact with the floor
Initially, as the ball is dropped from a height of 4m, it has a potential energy and no initial velocity, so its initial velocity is 0 m/s. The final velocity is the velocity just before the ball rebounds from the floor, which can be determined using the conservation of energy.
The total mechanical energy of the ball is conserved, so:
\[PE_i + KE_i = PE_f + KE_f\]
Where:
- \(PE_i\) is the initial potential energy
- \(KE_i\) is the initial kinetic energy (which is 0)
- \(PE_f\) is the final potential energy
- \(KE_f\) is the final kinetic energy (which is 0, as it momentarily comes to rest)
Since the ball's height changes from 4m to 3m, we can calculate the initial potential energy and the final potential energy using:
\[PE_i = mgh\]
\[PE_f = mgh\]
Where:
- \(m\) is the mass of the ball (which is not given)
- \(g\) is the acceleration due to gravity (approximately 9.8 m/s²)
- \(h\) is the height
Since the mass of the ball cancels out in the conservation of energy equation, we can use this equation to find the final velocity just before it rebounds:
\[0 + 0 = mgh + 0\]
\[mgh = mgh\]
\[0 = \frac{1}{2}mv_f^2 - mgh\]
Simplifying the equation, we get:
\[\frac{1}{2}mv_f^2 = mgh\]
\[v_f^2 = 2gh\]
\[v_f = \sqrt{2gh}\]
Now, let's calculate the value of \(v_f\):
\[v_f = \sqrt{2 \cdot 9.8 \cdot 4} = \sqrt{78.4} \approx 8.85 \, \text{m/s}\]
Given that the time of contact with the floor is 0.01 seconds, we can rearrange the kinematic equation to solve for acceleration:
\[a = \frac{{v_f - v_i}}{t}\]
Substituting the values:
\[a = \frac{{8.85 - 0}}{0.01} = 885 \, \text{m/s²}\]
Therefore, the magnitude of the acceleration during the contact is 885 m/s².
Since the ball is in contact with the floor, the acceleration would be directed upwards due to the upward force exerted by the floor. So, the direction of the acceleration is upwards.
h = 0.5g*t^2 = 4 m.
4.9*t^2 = 4
t^2 = 0.816
Tf = 0.904 s. = Fall time.
V1^2 = Vo^2 + 2g*h = 0 + 19.6*4 = 78.4
V1 = 8.85 m/s., Downward.
V^2 = V2^2 + 2g*h = 0
V2^2 = -2g*h = -2*(-9.8)*3 = 58.8
V2 = 7.67 m/s, Upward.
a = (V2-V1)/t = (7.67-8.85)/0.01 = -118
m/s^2, Upward.