Home
Class 11
PHYSICS
The block of mass M moving on the frict...

The block of mass `M` moving on the frictionless horizontal surface collides with the spring constant `k` and compresses it by length `L` . The maximum momention of the block after collision is

A

`(kL^(2))/(2 M)`

B

`sqrt(Mk) L(c )`S

C

`(ML^(2))/(k)`

D

zero

Text Solution

Verified by Experts

The correct Answer is:
B
(b) `(1)/(2) M nu^(2) = (1)/(2) k L6(2) rArr nu = sqrt(k)/(M) . L`
Momentum `= M xx nu = M xx sqrt (k)/(M). L = sqrt(kM) . L`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • WAVES

    SUNIL BATRA (41 YEARS IITJEE PHYSICS)|Exercise JEE Main And Advanced|122 Videos

Similar Questions

Explore conceptually related problems

A block of mass 2 kg is moving on a frictionless horizontal surface with a velocity of 1m//s towards another block of equal mass kept at rest. The spring constant of the spring fixed at one end is 100 N//m. Find the rnaximum compression of the spring

A block of mass m moving at a speed v_0 compresses a spring of spring constant k as shown in the figure. Find (a) the maximum compression of spring and (b) speed of the block when the spring is compressed to half of maximum compression.

Knowledge Check

  • A block of mass m moving with velocity v_(0) on a smooth horizontal surface hits the spring of constant k as shown. Two maximum compression in spring is

    A
    `sqrt((2m)/(k))v_(0)`
    B
    `sqrt((m)/(k)).v_(0)`
    C
    `sqrt((m)/(k)).v_(0)`
    D
    `(m)/(2k).v_(0)`

  • A body of mass 5 kg moving with a speed of 1.5 m/s on a horizontal smooth surface collides with a nearly weightless spring of force constant k = 5 N/m. The maximum compression of the spring would be

    A
    0.5 m
    B
    0.15 m
    C
    1.5 m
    D
    0.12 m

  • A block of mass m is moving with a speed v on a horizontal rought surface and collides with a horizontal monted spring of spring constant k as shown in the figure .The coefficient of friction between the block and the floor is mu The maximum cobnpression of the spring is

    A
    `- (mu mg)/(k) + (1)/(k) sqrt((mu mg)^(2) + mkv^(2))`
    B
    ` (mu mg)/(k) + (1)/(k) sqrt((mu mg)^(2) + mkv^(2))`
    C
    `- (mu mg)/(k) + (1)/(k) sqrt((mu mg)^(2) - mkv^(2))`
    D
    `(mu mg)/(k) + (1)/(k) sqrt((mu mg)^(2) + mkv^(2))`

    • Similar Questions

    Explore conceptually related problems

    A 16kg block moving on a frictionless horizontal surface with a velocity of 4m//s compresses an ideal spring and comes to rest. If the force constant of the spring be 100N//m , then how much is the spring commpressed ?

    mass of 0.5 kg moving with a speed of 1.5 m//s on a horizontal smooth surface, collides with a nearly weightless spring of force constant k = 50 N/m. The maximum compression of the spring would be

    A mass of 0.5kg moving with a speed of 1.5 m/s on a horizontal smooth surface, collides with a nearly weightless spring of force constant k=50 N/m . The maximum compression of the spring would be

    A mass of 0.5 kg moving with a speed of 1.5 m//s on a horizontal smooth surface, collides with a nearly weightless spring of force constant k =50 N//m The maximum compression of the spring would be.

    Two blocks of masses m and 2m rest on a frictionless horizontal surface. They are connected by an ideal spring of relaxed length l and stiffness constant k . By means of a massless thread connecting the blocks the spring is held compressed to a length l//2 . the whole system is moving with speed v in a direction perpendicular to the length of the spring. the thread is then burnt. Answer the following questions in terms of l,km,v After burning the string the speed of centre of mass is :

    Home
    Profile