In present problem, there are no net external forces. the sand is slowly leaking downwards  from the whole. Consequently, normal  reaction acting vertically upwards also slowly decreases.
But it does not do any work because it is at right angle to trolley's motion. Thus, there is no change in speed of the trolley and it will continue among moving with a  uniform speed of 27 kmh^-1

When a spring is compressed or stretched, potential energy of the spring increase in both cases. This is because work is done by us in compression as well as stretching

Let us consider the mass of the molecule be m and that of wall be M. The wall remains at rest due to its large mass. Resolving momentum of the molecule along x-axis and y-axis, we get
The x-component of momentum of molecule
= mu cos \(\theta\) = - m 200 cos 30° = -100\(\sqrt { 3 } \)m

y-component of the molecule
= mu sin \(\theta\) = m \(\times\) 200 \(\times\) sin 30° = 100 m
Before collision: x-component of total momentum (wall + molecule)
= 0 + (-100\(\sqrt { 3 } \) m) = -100 \(\sqrt { 3 } \) m
y- component of momentum (wall + molecule) = 0 + 100 m = 100 m
After collision: x-component of the momentum (wall + molecule)
= 0 + m 200 cos 30° = 100 \(\sqrt { 3 } \) m
and y-component = 0 + m 100 sin 30° = 100 m
We find that momentum of the (molecule + wall) system is conserved. The wall has a recoil momentum such that momentum of the wall + momentum of outgoing molecule equals the momentum of the incoming molecule.
Initial kinetic energy \(\left( \frac { 1 }{ 2 } { mu }^{ 2 } \right) \) is the same as final K.E. \(\left( \frac { 1 }{ 2 } { mv }^{ 2 } \right) \)of the molecule as u = v = 200 m/s i.e., thus, the collision is elastic collision.

During the short interval of an elastic collision, total linear momentum is conserved.

The value is almost always less than one due to initial translational kinetic energy being lost to rotational kinetic energy, plastic deformation, and heat. It can be more than 1 if there is an energy gain during the collision from a chemical reaction, a reduction in rotational energy, or another internal energy decrease that contributes to the post-collision velocity.
\(\text { Coefficient of restitution }(e)=\frac{\mid \text { Relative velocity after collision } \mid}{\mid \text { Relative velocity before collision } \mid}\)

In physics, circular motion is a movement of an object along the circumference of a circle or rotation along a circular path. It can be uniform, with constant angular rate of rotation and constant speed, or non-uniform with a changing rate of rotation. The rotation around a fixed axis of a three-dimensional body involves circular motion of its parts. The equations of motion describe the movement of the center of mass of a body.

Here, \(m_1=1.2 \mathrm{~kg}, u_1=20 \mathrm{~cm} / \mathrm{s}, m_2=1.2 \mathrm{~kg}, u_2=0\)
If v₁ and v₂ are velocities of the two blocks after collision, then accordin to the principle of conservation of momentum, \(m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2\)
\(1.2 \times 20+0=1.2 v_1+1.2 v_2 \therefore v_1+v_2=20(\mathrm{~cm} / \mathrm{s}) \ldots(\mathrm{i})\)
velocity of approach \(=u_1-u_2=20 \mathrm{~cm} / \mathrm{s}\), velocity of separation \(=v_2-v_1\)
By definition, \(e=\frac{v_2-v_1}{u_1-u_2}\)
\(\frac{3}{5}=\frac{v_2-v_1}{20} \therefore v_2-v_1=\frac{20 \times 3}{5}=12 \text {.. }\)
From (i) and (ii), \(v_1=4 \mathrm{~cm} / \mathrm{s}, v_2=16 \mathrm{~cm} / \mathrm{s}\)
Loss in K.E.
\( =\frac{1}{2} m_1 u_1^2-\frac{1}{2}\left(m_1\right) v_1^2-\frac{1}{2} m_2 v_2^2=\frac{1}{2} \)
\( \times 1.2\left(\frac{20}{100}\right)^2-\frac{1}{2}(1.2)\left(\frac{4}{100}\right)^2-\frac{1}{2} \times 1.2\left(\frac{16}{100}\right)^2 \)
\( =2.4 \times 10^{-2}-0.096 \times 10^{-2}-1.536 \times 10^{-2}\)
\(7.7\times { 10 }^{ -3 }\)

Friction always opposes motion. Hence, work done by a body against friction cause loss in its kinetic energy

Here, m₁ = mass of neutron = m
m₂ = mass of target nucleus = M
u₁ = u and u₂ = 0
Now,  \({ v }_{ 2 }=\frac { { 2 }m_{ 1 } }{ { m }_{ 1 }{ +m }_{ 2 } } .{ u }_{ 1 }+\frac { { m }_{ 2 }-{ m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } .{ u }_{ 2 }\)  
 \(=\frac { 2m }{ m+M } .u+0=\frac { 2mu }{ m+M } \)  
Initial KE of mass m, \({ K }_{ 1 }=\frac { 1 }{ 2 } { m }_{ 1 }{ u }_{ 1 }^{ 2 }=\frac { 1 }{ 2 } m{ u }^{ 2 }\)  
Final KE of mass M, \({ K }_{ 2 }=\frac { 1 }{ 2 } { m }_{ 2 }{ u }_{ 2 }^{ 2 }=\frac { 1 }{ 2 } M{ \left( \frac { 2mu }{ m+M } \right) }^{ 2 }=\frac { 2M{ m }^{ 2 }{ u }^{ 2 } }{ { \left( m+M \right) }^{ 2 } } \)  
Fraction of the initial KE transferred,
\(f=\frac { { K }_{ 2 } }{ { K }_{ 1 } } =\frac { 2M{ m }^{ 2 }{ u }^{ 2 } }{ { \left( m+M \right) }^{ 2 } } \times \frac { 2 }{ m{ u }^{ 2 } } =\frac { 4mM }{ { \left( m+M \right) }^{ 2 } } \)    
For deuterium, M = 2m, therefore
\(f=\frac { 4m\times 2m }{ { \left( m+2m \right) }^{ 2 } } =\frac { 8 }{ 9 } \simeq 0.9\) 
About 90% of the neutron's energy is transferred to deuterium.