Question:
A ball of mass \(300\;g\) is dropped from a height \(10\;m\) above a sandy ground. On reaching the ground, it penetrates through a distance \(1.5\;m\) in sand and finally stops. The average resistance offered by the sand to oppose the motion is
Take \(g=10\;\text{m s}^{-2}\).
A ball of mass \(300\;g\) is dropped from a height \(10\;m\) above a sandy ground. On reaching the ground, it penetrates through a distance \(1.5\;m\) in sand and finally stops. The average resistance offered by the sand to oppose the motion is
Take \(g=10\;\text{m s}^{-2}\).
Take \(g=10\;\text{m s}^{-2}\).
Show Hint
When an object penetrates a surface and stops, use the work-energy theorem. Include both the work done by gravity and the resisting force during penetration.
Updated On: Jun 22, 2026
- \(35\;N\)
- \(23\;N\)
- \(34\;N\)
- \(28\;N\)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Convert mass into SI unit.
Given mass of ball is \[ m=300\;g \] \[ m=0.3\;kg \] Height from which ball is dropped: \[ h=10\;m \] Distance penetrated in sand: \[ s=1.5\;m \] Acceleration due to gravity: \[ g=10\;\text{m s}^{-2} \]
Step 2: Find kinetic energy when ball reaches the ground.
The potential energy lost before reaching the ground becomes kinetic energy.
\[ K.E.=mgh \] \[ K.E.=0.3\times 10\times 10 \] \[ K.E.=30\;J \]
Step 3: Apply work-energy principle during penetration in sand.
Let the average resistance offered by sand be \(R\).
During penetration, gravity does positive work and sand resistance does negative work.
Work done by gravity: \[ mg s=0.3\times 10\times 1.5 \] \[ =4.5\;J \] Work done by resistance: \[ -Rs=-1.5R \] Since the ball finally stops, final kinetic energy is zero.
Therefore, \[ 30+4.5-1.5R=0 \]
Step 4: Solve for \(R\).
\[ 34.5=1.5R \] \[ R=\frac{34.5}{1.5} \] \[ R=23\;N \]
Step 5: Final conclusion.
Hence, the average resistance offered by the sand is \[ \boxed{23\;N} \]
Given mass of ball is \[ m=300\;g \] \[ m=0.3\;kg \] Height from which ball is dropped: \[ h=10\;m \] Distance penetrated in sand: \[ s=1.5\;m \] Acceleration due to gravity: \[ g=10\;\text{m s}^{-2} \]
Step 2: Find kinetic energy when ball reaches the ground.
The potential energy lost before reaching the ground becomes kinetic energy.
\[ K.E.=mgh \] \[ K.E.=0.3\times 10\times 10 \] \[ K.E.=30\;J \]
Step 3: Apply work-energy principle during penetration in sand.
Let the average resistance offered by sand be \(R\).
During penetration, gravity does positive work and sand resistance does negative work.
Work done by gravity: \[ mg s=0.3\times 10\times 1.5 \] \[ =4.5\;J \] Work done by resistance: \[ -Rs=-1.5R \] Since the ball finally stops, final kinetic energy is zero.
Therefore, \[ 30+4.5-1.5R=0 \]
Step 4: Solve for \(R\).
\[ 34.5=1.5R \] \[ R=\frac{34.5}{1.5} \] \[ R=23\;N \]
Step 5: Final conclusion.
Hence, the average resistance offered by the sand is \[ \boxed{23\;N} \]
Was this answer helpful?
0
0
Top AP EAPCET Physics Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The radii of the two columns in a U - tube are r1 and r2. When a liquid of density p (angle of contact is $ 0{}^\circ $ ) is filled in it, the level difference of the liquid in the two arms is h. The surface tension of the liquid is: (g = acceleration due to gravity)
- A rigid metallic sphere is spinning around its own axis in the absence of external torque. If the temperature is raised, its volume increases by $9 \%$. The change in its angular speed is
View More Questions
Top AP EAPCET Work-energy theorem Questions
- The ball initially has a velocity of 5 m/s in the positive x-direction and some time later has a velocity of 7 m/s in the positive y-direction. If the work done by the ball is 48 joule, then the mass of the ball in kg is:
- A loaded bus and an unloaded bus are both moving with the same kinetic energy. The mass of the former is twice that of the later. Brakes are applied to both so as to exert equal retarding forces. If $S_1$ and $S_2$ are the distances covered by the two buses before coming to rest respectively, then:
- A block of mass \(2\,\text{kg}\), moving along the \(X\)-axis on a horizontal surface with a speed of \(4\,\text{ms}^{-1}\), enters a rough surface from \(x=0.5\,\text{m}\) to \(x=1.5\,\text{m}\). The retarding force on this rough surface is \[ F=-12x\ \text{N}. \] The speed of the block as it just crosses the rough surface is
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions