A player caught a cricket ball of mass 150 g moving at a speed of 20 m/s. If the catching process is completed in 0.1 s, the magnitude of force exerted by the ball on the hand of the player is:
To calculate the force exerted by the cricket ball on the hand of the player, we can use the impulse-momentum theorem. The theorem states that the impulse experienced by an object is equal to the change in its momentum. Mathematically, this can be expressed as:
\(F \cdot \Delta t = \Delta p\)
Where:
\(F\) is the force exerted.
\(\Delta t\) is the time over which the force is applied.
\(\Delta p\) is the change in momentum.
First, compute the initial and final momentum:
The initial momentum \(p_i = m \cdot v_i\), where \(m = 0.15 \text{ kg}\) (mass of the ball converted to kg) and \(v_i = 20 \text{ m/s}\). Thus, \(p_i = 0.15 \times 20 = 3 \text{ kg m/s}\).
The final momentum \(p_f = 0 \text{ kg m/s}\) since the ball is caught and brought to rest.
The change in momentum \(\Delta p = p_f - p_i = 0 - 3 = -3 \text{ kg m/s}\).
Considering only the magnitude (ignore the negative sign), the impulse is equal to \(3 \text{ N s}\).
Now, calculate the force using the formula:
\(F = \frac{\Delta p}{\Delta t}\)
Substitute the values:
\(F = \frac{3}{0.1} = 30 \text{ N}\)
Therefore, the magnitude of the force exerted by the ball on the hand of the player is 30 N.
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Approach Solution -2
The force exerted is:
\[ F = \frac{\Delta P}{\Delta t}. \]
Here: \[ \Delta P = m \cdot v - 0 = 150 \times 10^{-3} \cdot 20 \, \text{kg m/s}, \] \[ F = \frac{150 \times 10^{-3} \cdot 20}{0.1}. \]
