Question

An ideal thread (perfectly inelastic) of length 2r is fixed at origin o' and other end is connected to block of moss The block is initially at rest at (10). An impulse 'J' is imported to block in direction at t-o. The block in its subsequent motion passes through y axis for the first time at t=​

Answer 1

Answer:

answer is 3

Explanation:

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One end of an ideal spring is fixed to a wall at origin O and axis of spring is parallel to x−axis. A block of mass m=1 kg is attached to free end of the spring and it is performing SHM. Equation of position of the block in coordinate system shown in figure is x=10+3sin10t. Here, t is in second and x in cm. Another block of mass M=3 kg, moving towards the origin with velocity 30 cm/s collides with the block performing SHM at t=0 and gets stuck to it calculate.

(i) New amplitude of oscillation

(ii) New equation for position of the combined body.

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Solution

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Correct option is

C

3

From equation x=10+3sin(10t)

So ω=10 and A=3

We know ω=

m

k

k=10

2

×1=100N/m

The velocity of mass m at t=0 is v

1

=ω×A=10×3=30cm/s

Now from momentum conservation

(M+m)v=30×M−v

1

×m

4v=30×3−30×1

v=15cm/s

Now to find the new amplitude of system

Let the new amplitude be A

1

By conservation of energy

2

(M+m)v

2

=

2

kA

1

2

A

1

=0.03m=3cm

and new angular velocity of system ω

1

=

M+m

k

ω=

4

100

=5rad/s

So the equation is x

′

=10−3sin5t

x=3

Answer 2

The block in its subsequent motion passes through y axis for the first time at t=10+3sin (from equation x=10+3sin) (from equation x=10+3sin) (from equation (10t)As a result, A=3 and =10

• m k = m k= m k= m k= m k=
• ​1=100N/m k=10 2 k=10 2 k=10 2 k=10 2 k=10 2 k
• At t=0, the velocity of mass m is v 1 =A=103=30cm/s.
• Now, based on momentum conservation (M+m)v=30Mv 1 m,
• 4v=30×3−30×1\sv=15cm/s
• Now it's time to determine the system's new amplitude.
• Let A 1 be the new amplitude.

2 (M+m)v 2 (M+m)v 2 (M+m)v 2 (M+m)v 2 (M

2 kA 2 kA 2 kA 2 kA 2 kA 2

​

A 1 =0.03m=3cm and a new system angular velocity of 1 = M+m k4 100 = 5 rad/s

As a result, the equation is x ′ =103sin5t x=3