(B) Let the retardation produced by the wooden block be $a$. Using the equation of motion $v^2 = u^2 + 2as$:For the first $24\,cm$ $(s_1 = 24\,cm)$:$(

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(B) Let the retardation produced by the wooden block be $a$. Using the equation of motion $v^2 = u^2 + 2as$:For the first $24\,cm$ $(s_1 = 24\,cm)$:$(\frac{u}{3})^2 = u^2 - 2a(24)$$\frac{u^2}{9} = u^2 - 48a$$48a = u^2 - \frac{u^2}{9} = \frac{8u^2}{9}$$a = \frac{8u^2}{9 \times 48} = \frac{u^2}{54}.........(1)$Now,let the total length of the block be $L$. For the entire motion until the bullet comes to rest $(v=0)$:$0^2 = u^2 - 2aL$$2aL = u^2$$L = \frac{u^2}{2a}.........(2)$Substituting the value of $a$ from $(1)$ into $(2)$:$L = \frac{u^2}{2 \times (u^2/54)} = \frac{54}{2} = 27\,cm$.

Class 11 Physics Motion in Straight Line Uniformly Accelerated Motion

Medium

$A$ bullet from a gun is fired into a rectangular wooden block with velocity $u$. When the bullet travels $24\,cm$ through the block along its length horizontally,the velocity of the bullet becomes $\frac{u}{3}$. It then further penetrates into the block in the same direction before coming to rest exactly at the other end of the block. The total length of the block is $........\,cm$.

NEET 2023 All NEET

A
$30$
B
$27$
C
$24$
D
$28$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

Motion in Straight Line

Subtopic

Uniformly Accelerated Motion

Previous Year

NEET 2023 Paper All NEET years →

Two cars $P$ and $Q$ are moving on a road in the same direction. The acceleration of car $P$ increases linearly with time,whereas car $Q$ moves with a constant acceleration. Both cars cross each other at time $t = 0$ for the first time. The maximum possible number of crossings (including the crossing at $t = 0$) is . . . . . . .

DifficultJEE MAIN 2025

View Solution

$A$ body of mass $10 \,kg$ has a constant velocity of $10 \,m \,s^{-1}$. $A$ constant force is applied for a duration of $4 \,s$ on the object, which gives it a speed of $2 \,m \,s^{-1}$ in the opposite direction. Find the acceleration produced in it.

MediumAP EAMCET 2020

View Solution

$A$ particle moves according to the equation $y = a + bt + ct^2 - dt^4$. What are the initial velocity and initial acceleration of the particle?

Medium

View Solution

$A$ particle starts with an initial velocity of $10.0 \, m/s$ along the $x$-direction and accelerates uniformly at the rate of $2.0 \, m/s^2$. The time taken by the particle to reach the velocity of $60.0 \, m/s$ is $....... \, s$.

EasyJEE MAIN 2023

View Solution

An automobile travelling with a speed of $60\,km/h$ can brake to stop within a distance of $20\,m$. If the car is going twice as fast,i.e.,$120\,km/h$,the stopping distance will be ........... $m$.

MediumAIEEE 2004

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Similar Questions

$A$ body of mass $10 \,kg$ has a constant velocity of $10 \,m \,s^{-1}$. $A$ constant force is applied for a duration of $4 \,s$ on the object, which gives it a speed of $2 \,m \,s^{-1}$ in the opposite direction. Find the acceleration produced in it.

$A$ particle moves according to the equation $y = a + bt + ct^2 - dt^4$. What are the initial velocity and initial acceleration of the particle?

$A$ particle starts with an initial velocity of $10.0 \, m/s$ along the $x$-direction and accelerates uniformly at the rate of $2.0 \, m/s^2$. The time taken by the particle to reach the velocity of $60.0 \, m/s$ is $....... \, s$.

An automobile travelling with a speed of $60\,km/h$ can brake to stop within a distance of $20\,m$. If the car is going twice as fast,i.e.,$120\,km/h$,the stopping distance will be ........... $m$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module