Question 334: Work, Energy and Power & Circular Motion

Question

A spring-block system is resting on a frictionless floor as shown in the figure. The spring constant is $2.0\,N{m^{ - 1}}$ and the mass of the block is $2.0$ $kg.$ Ignore the mass of the spring. Initially the spring is an unstretched condition. Another block of mass $1.0$ $kg$ moving with a speed of $2.0$ $m{s^{ - 1}}$ collides elastically with the first block. The collision is  such that the $2.0$ $kg$ block does not hit the wall. The distance,  in metres, between the two blocks when the spring returns to its unstretched position for the first time after the collision is __________.

Accepted Answer

After collision the 2.0 kg block will perform simple harmonic oscillation with time period
$T = 2\pi \sqrt {{m \over k}} $
Given m = 2.0 kg and k = 2.0 N m
$-$1
, we have
T = 2$\pi$
Thus, the block returns to its original position in time
$t = {T \over 2}$ = $\pi$ s
That is, t = 3.14 s
Now, if v
1
and v
2
are velocities of 1.0 kg block and 2.0 kg block, respectively, before collision; v'
1
and v'
2
are velocities of 1.0 kg block and 2.0 block, respectively, after collision. So, by conservation of momentum
m
1
v
1
+ m
2
v
2
= m
1
v'
1
+ m
2
v'
2
Here, m
1
= 1.0 kg, v
1
is initial speed of 1.0 kg block = 2.0 m s
$-$1
, m
2
= 2.0 kg, v
2
is initial speed of 2.0 kg block = 0.0 m s
$-$1
, v'
1
is final speed of 1.0 kg block after collision and v'
2
is final speed of 2.0 kg block after collision. Then, 1.0 kg $	imes$ 2.0 m/s + 2.0 kg $	imes$ 0 m/s = 1.0 v'
1
+ 2.0 v'
2
v'
1
+ 2v'
2
= 2 ..... (1)
Also, using definition of coefficient of restitution
v'
2
$-$ v'
1
= $\varepsilon $(v
1
$-$ v
2
)
Since collision is elastic, So $\varepsilon $ = 1
$\Rightarrow$ v'
2
$-$ v'
1
= v
1
$-$ v
2
$\Rightarrow$ v'
2
$-$ v'
1
= 2 $-$ 0
$\Rightarrow$ v'
2
$-$ v'
1
= 2 ...... (2)
From Eqs. (1) and (2), we get
$v{'_2} = {4 \over 3}$ m s
$-$1
and $v{'_1} = {-2 \over 3}$ m s
$-$1
Therefore, distance between the blocks is given as
$s = v{'_1} 	imes t = {{ - 2} \over 3} 	imes 3.14 = 2.09$ m

Work, Energy and Power & Circular Motion

Integer / Numeric Answer Type