Question

: A 500-g mass rests freely atop a spring (i.e.-isn

Answer 1

i) The initial compression of the string is unaffected by the 200-g mass, only puts the 500-g mass in equilibrium:

Hooke's law says that the spring restoring force is: $\left|F_s\right|=kx$ with  x being the compression or stretching. Since it is compressed downwards, it goes upwards to balance the weight of the 500-g mass:

$kx=m_{B}\cdot g \\ (400 N/m)\cdot x=(0.500 kg)\cdot (9.81 m/s^2)\Rightarrow x=0.0123 m=1.23 cm$

ii)We will denote "b.c. = (just) before collision" and "a.c.= (just) after collision". We first need to know with what velocity the 200-g (block A) mass reaches the other block JUST before impact. For that we use conservation of mechanical energy, since only the conservative force of gravity acts on A:

we use conservation of mechanical energy for only the mass A here, measuring height from the bottom where the other block is:

where K is kinetic energy, Ug is gratitational potential energy and Uelast is elastic potential energy. Initially, K is zero, at the end Ug is zero and Uelast is always zero (no spring being considered as part of the system of the mass A alone):

$E_{A, initial}=E_{A, final}\\ 0+mgh+0=\frac{1}{2}mv_{A, b.c.}^2+0+0\Rightarrow v_{A, b.c.}=-\sqrt{2gh}$

where we take the minus sign when taking the square rooot because it's going downwards.

Now, we consider the collision. During collisions, the duration is short in time and no net force is to be considered. This is tantamount to considering conservation of momentum for the system of two masses:

$p_{system, b.c.}=p_{system, a.c.}\\ p_{A, b.c.}+p_{B, b.c.}=p_{A, a.c.}+p_{B, a.c.}\\ m_A \cdot (-\sqrt{2gh})+0=m_A v_{A, a.c.}+m_B v_{B, a.c.}\\- \sqrt{2gh}= v_{A, a.c.}+\frac{m_B}{m_A} v_{B, a.c.} \dots (I)$

Usually, mechanical energy is not conserved in collisions, but it is conserved only if the collision is ELASTIC (an ideal case) and this is what we are told to consider. However, writing down conservation of mechanical energy is cumbersome, we instead use the fact that it leads to a restitution coefficient of 1:

$-\frac{v_{A, a.c.}-v_{B, a.c.}}{v_{A, b.c.}-v_{B, b.c.}}=1\Rightarrow v_{A, a.c.}-v_{B, a.c.}=v_{B, b.c.}-v_{A, b.c.}$

$v_{A, a.c.}-v_{B, a.c.}=-\left (-\sqrt{2gh} \right )\\\sqrt{2gh}=v_{A, a.c.}-v_{B, a.c.}\dots (II)$   We need to solve the system of (I) and (II).

We can subtract (II) from (I) and get: $2\sqrt{2gh}=-\left(1+\frac{m_B}{m_A}\right)v_{B, a.c.}\Rightarrow v_{B, a.c.}=-\left (\frac{2\sqrt{2gh}}{\frac{m_B}{m_A}+1} \right )$

and substituting this back into (II): $v_{A, a.c.}=\left (\frac{\frac{m_B}{m_A}-1}{1+\frac{m_B}{m_A}} \right )\sqrt{2gh}$ . this is the velocity of the mass A (the 200-g mas) just after collision, the bounce velocity, which is positive and means it will reach a maximum height upwards.

Again, we use conservation of mechanical energy as we did at the beginning to find the maximum height of A.

$E_{A, initial}=E_{A, final}\\ \frac{1}{2}m_A v_{A, a.c.}^2+0+0=0+m_A g h_{max}+0\Rightarrow h_{max}= \frac{v_{A, a.c.}^2}{2g}$ and using what we found for the velocity after collision we get: $h_{max}=\left (\frac{\frac{m_B}{m_A}-1}{\frac{m_B}{m_A}+1} \right )^2\cdot h$ . Now we plug in values:

$h_{max}=0.551m$

(iii)WE will use the velocity of the mass B just after collision we found earlier: $v_{B, a.c.}=-\left (\frac{2\sqrt{2gh}}{\frac{m_B}{m_A}+1} \right )$

We also need the IMPORTANT assumption we are told to make: That during collision any motion is to be neglected. We can consider that the spring only deforms just after collision to use energies correctly. This is important and was remarked.

Now, we consider the system "Mass B+spring". There are only conservative external forces: gravity and the potential energy of the spring. Conservation of mechanical energy is again used. Of course we assume the spring is massless:

Here, the reference heights will be measured wrt the initial height of the mass, therefore, final height willbe negative (below). This means that and are both 0 and    $U_{elast, fin}=\frac{1}{2}k (-x)^2$ . Also, at the end the mass is at that instant at rest and $U_g_{B, fin} =mg\cdot (-x)$ .

$\frac{1}{2}m_Bv_{B, a.c.}^2+0 +0=0-m_Bgx +\frac{1}{2}k x^2$ and substituting back the result $v_{B, a.c.}=-\left (\frac{2\sqrt{2gh}}{\frac{m_B}{m_A}+1} \right )$ we get:

$\frac{1}{2}k x^2-m_Bgx -\left (\frac{4m_Bgh}{\frac{m_B}{m_A}+1} \right )=0$

We need to solve this quadratic equation: $x=\frac{m_Bg+\sqrt{m_B^2g^2+\frac{8m_Bghk}{\frac{m_B}{m_A}+1}}}{k}$ .

The value is now: $x=0.302m$

(iv) For this we can use again conservation of mechanical energy FROM the moment the mass B starts deforming the spring downwards (as in part iii) until it reaches the maximum height, that's because what happens it between doesn't matter, mechanical energy is conserved for the system, and the spring is not attached, that means the spring recovers and gives all the energy back to the mass. We only focus on the mass B:

$E_{B, initial}=E_{B, final}\\ K_{B, a.c.}+U_g_{B, ini} =K_{B, fin}+U_g_{B, fin} \\ \frac{1}{2}m_Bv_{B, a.c.}^2+0=0+m_Bg \cdot h_{max}\\ \left (\frac{4m_Bgh}{\frac{m_B}{m_A}+1}\right)=m_bgh_{max}\\$

$h_{max}=\frac{4h}{\frac{m_B}{m_A}+1}=3.43m$