56 PE
Question
A steel ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.45 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 0.0800 ms (\({\bf{8}}{\bf{.00 \times 1}}{{\bf{0}}^{{\bf{ - 5}}}}\;{\bf{s}}\)) . (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?
Step-by-Step Solution
Verified(a) The strike velocity of ball with floor is 5.42 m/s.
(b) The velocity of ball just after strike with floor is 5.33 m/s.
(c) The acceleration of ball during contact with floor is \({\bf{1}}{\bf{.34 \times 1}}{{\bf{0}}^{\bf{5}}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\).
(d) The compression of ball during strike is \({\bf{1}}{\bf{.1 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}\;{\bf{m}}\).
The initial height of ball from floor is \(H = 1.50\;{\rm{m}}\).
The rebound height of ball from floor is \(h = 1.45\;{\rm{m}}\).
The contact duration of ball with floor is \(t = 8 \times {10^{ - 5}}\;{\rm{s}}\)
The above problem is based on the collision principle and velocity of ball just before strike with floor is found by energy conservation.
The strike velocity of ball with floor is given as
\(u = \sqrt {2gH} \)
Here, \(g\) is the gravitational acceleration and its value is \(9.80\;{\rm{m}}/{{\rm{s}}^2}\).
Substitute all the values in the above equation.
\(\begin{array}{l}u = \sqrt {2\left( {9.80\;{\rm{m}}/{{\rm{s}}^2}} \right)\left( {1.50\;{\rm{m}}} \right)} \\u = 5.42\;{\rm{m}}/{\rm{s}}\end{array}\)
Therefore, the strike velocity of ball with floor is \(5.42\;{\rm{m}}/{\rm{s}}\).
The velocity of ball just after strike with floor is given as:
\(v = \sqrt {2gh} \)
Here, \(g\) is the gravitational acceleration and its value is \(9.80\;{\rm{m}}/{{\rm{s}}^2}\).
Substitute all the values in the above equation.
\(\begin{array}{l}v = \sqrt {2\left( {9.80\;{\rm{m}}/{{\rm{s}}^2}} \right)\left( {1.45\;{\rm{m}}} \right)} \\v = 5.33\;{\rm{m}}/{\rm{s}}\end{array}\)
Therefore, the velocity of ball just after strike with floor is \(5.33\;{\rm{m}}/{\rm{s}}\) .
The acceleration of ball during contact with floor is given as
\(a = \frac{{u + v}}{t}\)
Substitute all the values in the above equation.
\(\begin{array}{l}a = \frac{{5.42\;{\rm{m}}/{\rm{s}} + 5.33\;{\rm{m}}/{\rm{s}}}}{{8 \times {{10}^{ - 5}}\;{\rm{s}}}}\\a = 1.34 \times {10^5}\;{\rm{m}}/{{\rm{s}}^2}\end{array}\)
Therefore, the acceleration of ball during contact with floor is \(1.34 \times {10^5}\;{\rm{m}}/{{\rm{s}}^2}\).
The compression of ball during strike is given as
\(x = \frac{{{u^2}}}{{2a}}\)
Substitute all the values in the above equation.
\(\begin{array}{l}x = \frac{{{{\left( {5.42\;{\rm{m}}/{\rm{s}}} \right)}^2}}}{{2\left( {1.34 \times {{10}^5}\;{\rm{m}}/{{\rm{s}}^2}} \right)}}\\x = 1.1 \times {10^{ - 4}}\;{\rm{m}}\end{array}\)
Therefore, the compression of ball during strike is \(1.1 \times {10^{ - 4}}\;{\rm{m}}\).