In the figure, block 2 of mass 1.80 kg oscillates on the end of a spring in SHM

Question

In the figure, block 2 of mass 1.80 kg oscillates on the end of a spring in SHM with a period of 16.00 ms. The position of the block is given by                     x = (1.60 cm) cos(?t + ?/2). Block 1 of mass 3.60 kg slides toward block 2 with a velocity of magnitude 5.00 m/s,                    directed along the spring's length. The two blocks undergo a completely inelastic collision at time t = 4.00 ms. (The duration of the                    collision is much less than the period of motion.) What is the amplitude of the SHM after the collision? Please use my numbers.

Explanation / Answer

after collision both of them stuck so, new w' = sqrt(k/(m_1+m_2)

and at that time

X(t) = A cos(w't+theta)

V(t) = -Aw' sin(w't+theta)

Now coming back to what has happened before collision

w=2*pi*sqrt(k/m2) so, time period=sqrt(m2/k) so, k= m2/t^2=112500 N/m

V of m1 before collision=5 m/s;

V of m2 before collision=-1.6*w*sin(w*t+pi/2)=2*pi=0m/s;

now conserving moment

P = m1 *5 + m2 *0 = (m1+m2) V so, V= 5/5.4m/s

extension of spring at that point=1.6*cos(w*t+pi/2)=-1.6 cm=X

If you want to solve using energy (energy is not conserved in the collision), you can use

U = (1/2) k X^2

but k =m_2 w^2

So

(1/2) k A^2 = (1/2) k X^2 - (1/2)(m_1+m_2) V^2 (energy is applied in opp direction)

so, A=0.01465 m

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