Tags Laws of Motion

Laws of Motion

Get insights from 109 questions on Laws of Motion, answered by students, alumni, and experts. You may also ask and answer any question you like about Laws of Motion

Follow Ask Question

109

Questions

Discussions

Active Users

Followers

New answer posted

2 months ago

Laws of Motion Class 11th

Two masses A and B, each of mass M are fixed together by a massless spring. A force acts on the mass B as shown in figure. If the mass A starts moving away from mass B with acceleration 'a', then the acceleration of mass B will be:

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

Let the spring is in extension state and $a_{B} > a_{A}, {\vec{a}}_{A B} = - a_{A} \hat{i} - (- a_{B} \hat{i}) = (- a_{A} + a_{B}) \hat{i}$

Hence we can say that block moves away from block B in the frame of B

F – kx = Ma_B . (1)

kx = Ma . (2)

$\Rightarrow a_{B} = \frac{F}{M} - a$

0 0

New answer posted

2 months ago

Laws of Motion Class 11th

Kindly go through the questions

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

0 0

New answer posted

2 months ago

Laws of Motion Class 11th

A car starts from rest and accelerates at 5 m/s². At t = 4 s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at
t = 6 s ? (Take g = 10 m/s² )

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

velocity of car at t = 4sec is
v = u + at
v = 0 + 5 (4)
= 20 m/s
At t = 6sec
acceleration is due to gravity ∴ a = g = 10 m/s
v? = 20 m/s (due to car)
v? = u + at
= 0 + g (2) (downward)
= 20 m/s (downward)
v = √ (v? ² + v? ²)
= √ (20² + 20²)
= 20√2 m/s

0 0

New answer posted

2 months ago

Laws of Motion Class 11th

Calculate the maximum acceleration of a moving car so that a body lying on the floor of the car remains stationary. The coefficient of static friction between the body and the floor is $0.15 (g = 10 {m s}^{- 2})$ .

0 Follower 8 Views

alok kumar singh

Contributor-Level 10

$R = R_{0} (1 + α Δ T)$

$\begin{array}{r} \Rightarrow 6.8 = 2 [1 + α * (80 - 0)] \\ \Rightarrow α = \frac{3.4 - 1}{80} \\ = 0.03 \\ = 3 * 10^{- 2}^{?} C^{- 1} \end{array}$

0 0

New answer posted

2 months ago

Laws of Motion Class 11th

A 10μF capacitor is fully charged to a potential difference of 50 V. After removing the source voltage, it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes 20 V. The capacitance of the second capacitor is:

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

By conservation of charge, (50) (10) = 20 (10+C) ⇒ C=15µF

0 0

New answer posted

2 months ago

Laws of Motion Class 11th

A block of mass m = 1 kg slides with velocity v = 6 m/s on a frictionless horizontal surface and collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about O and swings as a result of the collision making angle θ before momentarily coming to rest. If the rod has mass M = 2 kg, and length l = 1 m, the value of θ is approximately: (take g = 10 m/s²)

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

By Conservation of Angular Momentum (COAM) about O, just before and after the collision:
Initial angular momentum L? = mv?
Final angular momentum L? = Iω? = (I_rod + I_block)ω? = (M? ²/3 + m? ²)ω?
L? = L?
mv? = (M? ²/3 + m? ²)ω?
ω? = mv / (M? /3 + m? ) = (16) / (21/3 + 1*1) = 6 / (5/3) = 18/5 rad/s

By Conservation of Total Mechanical Energy (COTME) after collision until it comes to rest:
Initial Energy (at the lowest point) = Rotational K.E. = (1/2)Iω? ²
Final Energy (at the highest point) = Potential Energy = mgh_m + Mg h_M
mgh_m = mg (? -? cosθ)
Mg h_M = Mg (? /2 - (? /2)cosθ)
(1/2) * (M? ²/3 + m? ²) * ω? ² = (mg + Mg/2) *

...more

0 0

New answer posted

2 months ago

Laws of Motion Class 11th

A particle of mass m with an initial velocity ui collides perfectly elastically with a mass 3m at rest. It moves with a velocity vj after collision, then v is given by

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

From momentum conservation
mui + 0 = mvj + 3mv'
v' = u/3 I - v/3 j
From kinetic energy conservation 1/2 mu² = 1/2 mv² + 1/2 (3m) (u/3)²+ (v/3)²)
Solving, v = u/√2

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
679k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Laws of Motion

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience