A cricket player catches a ball of mass 120 g moving with 25 m/s speed. If the catching process is completed in 0.1 s then the magnitude of force exerted by the ball on the hand of player will be (in SI unit):
- 24
- 12
- 25
- 30
The Correct Option is D
Approach Solution - 1
To solve the problem of determining the force exerted by the ball on the player's hand, we can use the impulse-momentum theorem. The impulse experienced by an object is equal to the change in momentum of the object. The formula we use is:
\(F \cdot \Delta t = \Delta p\)
- Where \(F\) is the force,
- \(\Delta t\) is the time duration (0.1 s in this case), and
- \(\Delta p\) is the change in momentum.
The change in momentum \(\Delta p\) is given by:
\(\Delta p = m(v_f - v_i)\)
- Where \(m\) is the mass of the ball (120 g = 0.12 kg),
- \(v_f\) is the final velocity (0 m/s, as the ball comes to rest), and
- \(v_i\) is the initial velocity (25 m/s).
Substituting the values:
\(\Delta p = 0.12 \cdot (0 - 25) = -3 \, \text{kg} \cdot \text{m/s}\)
According to the impulse-momentum theorem:
\(F \cdot 0.1 = -3\)
Solve for \(F\):
\(F = \frac{-3}{0.1} = -30 \, \text{N}\)
The negative sign indicates the force is in the opposite direction of the initial motion, but since we are asked for the magnitude, it is 30 N. Therefore, the correct answer is:
30
Approach Solution -2
Given: - Mass of the ball: \( m = 120 \, \text{g} = 0.12 \, \text{kg} \) - Initial speed of the ball: \( v = 25 \, \text{m/s} \) - Time taken to catch the ball: \( t = 0.1 \, \text{s} \) - Final speed of the ball: \( v_f = 0 \, \text{m/s} \) (since the ball is caught and comes to rest)
Step 1: Calculating the Change in Momentum
The change in momentum (\( \Delta p \)) of the ball is given by:
\[ \Delta p = m \cdot (v_f - v) \]
Substituting the given values:
\[ \Delta p = 0.12 \cdot (0 - 25) \, \text{kg} \cdot \text{m/s} \] \[ \Delta p = -3 \, \text{kg} \cdot \text{m/s} \]
The negative sign indicates a decrease in momentum.
Step 2: Calculating the Force Exerted
The force exerted by the ball on the hand of the player is given by Newton’s second law:
\[ F = \frac{\Delta p}{t} \]
Substituting the values:
\[ F = \frac{-3}{0.1} \, \text{N} \] \[ F = -30 \, \text{N} \]
The magnitude of the force is:
\[ |F| = 30 \, \text{N} \]
Conclusion:
The magnitude of the force exerted by the ball on the hand of the player is \( 30 \, \text{N} \).
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main laws of motion Questions
- In the given arrangement of a doubly inclined plane, two blocks of masses \( M \) and \( m \) are placed. The blocks are connected by a light string passing over an ideal pulley as shown. The coefficient of friction between the surface of the plane and the blocks is 0.25. The value of \( m \), for which \( M = 10 \, \text{kg} \) will move down with an acceleration of \( 2 \, \text{m/s}^2 \), is: (take \( g = 10 \, \text{m/s}^2 \) and \( \tan 37^\circ = 3/4 \))
- Three blocks M1, M2, M3 having masses 4 kg, 6 kg and 10 kg respectively are hanging from a smooth pully using rope 1, 2 and 3 as shown in figure. The tension in the rope 1, T1 when they are moving upward with acceleration of 2ms–2 is ............... N (if g = 10 m/s2 )
- A light unstretchable string passing over a smooth light pulley connects two blocks of masses \(m_1\) and \(m_2\). If the acceleration of the system is \(\frac{g}{8}\), then the ratio of the masses \(\frac{m_2}{m_1}\) is:
The driver sitting inside a parked car is watching vehicles approaching from behind with the help of his side view mirror, which is a convex mirror with radius of curvature \( R = 2 \, \text{m} \). Another car approaches him from behind with a uniform speed of 90 km/hr. When the car is at a distance of 24 m from him, the magnitude of the acceleration of the image of the side view mirror is \( a \). The value of \( 100a \) is _____________ m/s\(^2\).
- A body of mass 2 kg moving with velocity of $ \vec{v}_{\text{in}} = 3 \hat{i} + 4 \hat{j} \, \text{ms}^{-1} $ enters into a constant force field of 6N directed along positive z-axis. If the body remains in the field for a period of $ \frac{5}{3} $ seconds, then velocity of the body when it emerges from force field is:
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.