Another

In the Fig , block 2 of mass 2.0 2.0 kg oscillates on the end of a spring in SHM with a period of 20 20 ms. The position of the block is given by x ( t ) = ( 1.0 c m ) c o s ( ω t + π / 2 ) x (t )= (1.0 cm)\ cos (\omega t + \pi/2) . Block 1 1 of mass 4.0 4.0 kg slides toward block 2 2 with a velocity of magnitude 6.0 6.0 m/s, directed along the spring’s length.The two blocks undergo a completely inelastic collision at time t = 5.0 t = 5.0 ms. (the duration of the collision is much less than the period of motion.) What is the amplitude of the SHM after the collision?


The answer is 0.024.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

The question should specify the final unit of the answer

Yes same mistake

rajdeep brahma - 3 years, 4 months ago
Matteo De Zorzi
Mar 24, 2015

By the conservation of the linear momentum you can derive the velocity after the collison. After the collision the sum of the potential energy and kinect energy must be equal to 1/2kA^2. Derive the amplitude and that's all.

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...