Dynamics

The mass is initially at rest, so the total momentum is zero before and after the explosion. The two parts therefore have equal and opposite momenta: $p_1=p_2=p$.

Kinetic energy in terms of momentum: $K=\dfrac{p^2}{2m}$, so $K\propto\dfrac1m$ for equal momentum.

$\frac{K_{\text{small}}}{K_{\text{large}}}=\frac{m_{\text{large}}}{m_{\text{small}}}=\frac{40}{4}=10$

$K_{\text{small}}=10\times10\ \text{J}=100\ \text{J}$

The smaller mass has an initial kinetic energy of $100$ J.

Angle of friction ($\lambda$): the angle between the normal reaction $R$ and the resultant of the normal reaction and the limiting friction $F$.

Angle of repose ($\alpha$): the minimum angle of inclination of a plane to the horizontal at which a body placed on it just begins to slide down.

Relation. Since the resultant of $R$ and $F$ makes angle $\lambda$ with $R$,

$\tan\lambda=\frac{F}{R}=\mu$

Now take a body of weight $mg$ on a plane inclined at $\alpha$, just on the point of sliding. Along the plane, $F=mg\sin\alpha$. Perpendicular to the plane, $R=mg\cos\alpha$. Dividing,

$\frac{F}{R}=\tan\alpha=\mu$

Comparing the two results, $\tan\lambda=\tan\alpha$, so

$\lambda=\alpha$

The angle of friction equals the angle of repose, and both give $\mu=\tan\lambda=\tan\alpha$.

Coefficient of friction: the ratio of the force of friction to the normal reaction between the two surfaces, $\mu=\dfrac{F}{R}$. It has no unit and depends on the nature of the surfaces. For limiting friction it is called the coefficient of limiting (static) friction, $\mu_s=\dfrac{F_{\max}}{R}$.

Proof that $\mu_s=\tan\lambda$: The angle of friction $\lambda$ is the angle between the normal reaction $R$ and the resultant $S$ of the normal reaction and the limiting friction $F$.

Let a body rest on a rough horizontal surface and be on the point of sliding. The surface exerts a normal reaction $R$ upwards and limiting friction $F$ along the surface, opposing the tendency to move. The resultant $S$ of $R$ and $F$ makes angle $\lambda$ with $R$. Since $R$ and $F$ are perpendicular, from the right-angled triangle of forces,

$\tan\lambda=\frac{F}{R}$

But by definition $\dfrac{F}{R}=\mu_s$, hence

$\mu_s=\tan\lambda$

Take the initial direction of X as positive. Then $m_X=0.1$ kg, $u_X=6$ m/s, $m_Y=0.2$ kg, $u_Y=0$, $v_X=-2$ m/s.

By conservation of momentum,

$m_Xu_X+m_Yu_Y=m_Xv_X+m_Yv_Y$ $0.1(6)+0=0.1(-2)+0.2\,v_Y$ $0.6=-0.2+0.2\,v_Y\Rightarrow v_Y=\frac{0.8}{0.2}=4\ \text{m s}^{-1}$

Y moves with $4\ \text{m s}^{-1}$ in the original direction of X.

Elastic collision: a collision in which both the momentum and the kinetic energy of the system are conserved (no kinetic energy is converted to heat, sound or deformation).

Proof: Let masses $m_1,m_2$ move along a line with velocities $u_1,u_2$ before and $v_1,v_2$ after the collision.

Momentum conservation: $m_1(u_1-v_1)=m_2(v_2-u_2)\quad\text{...(1)}$

Kinetic energy conservation: $\tfrac12m_1u_1^2+\tfrac12m_2u_2^2=\tfrac12m_1v_1^2+\tfrac12m_2v_2^2$ $m_1(u_1^2-v_1^2)=m_2(v_2^2-u_2^2)\quad\text{...(2)}$

Divide (2) by (1): $u_1+v_1=v_2+u_2$ $\Rightarrow\ u_1-u_2=v_2-v_1$

The left side is the relative velocity of approach and the right side is the relative velocity of recession, so they are equal in magnitude.

(a) $ma$

Let the resistive force be $R$. Then $F-R=ma$ and $2F-R=3ma$. Subtracting the first from the second gives $F=2ma$. Then $R=F-ma=2ma-ma=ma$.

(c) linear momentum

A rocket expels hot gases backwards at high speed. As no external force acts on the rocket-plus-gas system, the total momentum stays constant, so the rocket gains an equal forward momentum.

(c) Linear momentum

In any collision the impulsive forces are internal, so the total linear momentum of the system is conserved. Kinetic energy is conserved only in elastic collisions, and angular momentum and torque depend on external conditions.

(a) constant velocity

By Newton's first law, a body moving with constant velocity has zero acceleration, so the net force on it is zero. Circular motion and any accelerated motion each need a net force.

(c) $5.8\ \text{m/s}^2$

Mass $m=\dfrac{25}{10}=2.5\ \text{kg}$ and friction $f=\mu N=0.3\times25=7.5\ \text{N}$.

$a=\frac{F-f}{m}=\frac{22-7.5}{2.5}=5.8\ \text{m/s}^2$

(a)

• First law: a body remains at rest or in uniform motion in a straight line unless an external net force acts on it.
• Second law: the rate of change of momentum is proportional to the applied force and takes place in the direction of the force, giving $F=ma$.
• Third law: to every action there is an equal and opposite reaction.

(b) Constant velocity means zero acceleration, so the applied force just balances friction:

$F=\mu mg=0.20\times10\times9.8=19.6\ \text{N}\approx20\ \text{N}$

(c) With the force removed, friction alone decelerates the body:

$a=\mu g=0.20\times9.8=1.96\ \text{m s}^{-2}$

$t=\frac{v}{a}=\frac{5}{1.96}\approx2.6\ \text{s}$

(c)

For a rigid body under coplanar forces to be in equilibrium, two conditions must hold together:

• Net force zero — the force vectors must form a closed polygon (head of the last vector meeting the tail of the first).
• Net torque zero — the lines of action of the forces must be concurrent (pass through a common point), or the moments must cancel.

Only diagram (c) satisfies both: its three force vectors close into a triangle, and their lines of action meet at a single point, so both the resultant force and the resultant torque vanish. In the other diagrams the vectors do not close (unbalanced force), or they form a couple (equal, opposite, parallel forces) that leaves a net torque.

(a) If no net external force acts on a system, the total linear momentum of the system remains constant. For two bodies,

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

(b) Given $m=4\times10^5\ \text{kg}$, $u=5000\ \text{m s}^{-1}$, $v=0$, $t=2\ \text{min}=120\ \text{s}$.

$F=\frac{m(v-u)}{t}=\frac{4\times10^5\,(0-5000)}{120}=-1.67\times10^{7}\ \text{N}$

The average force exerted by the ground on the aeroplane is $1.67\times10^{7}\ \text{N}$, opposite to its motion.

(c) Momentum is conserved for the aeroplane and the Earth together. The force of the ground on the aeroplane is internal to this system, and by Newton's third law the aeroplane exerts an equal and opposite force on the Earth. The Earth therefore gains the momentum the aeroplane loses, so the total momentum stays the same. Because the Earth's mass is enormous, its change in velocity is negligibly small.

Law of conservation of momentum: In the absence of an external resultant force, the total linear momentum of an isolated system remains constant. That is, the total momentum before an interaction is equal to the total momentum after the interaction.

For two bodies colliding,

$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$

where $u$ and $v$ are velocities before and after collision.

Given: Mass of jet, $m = 4\times10^5\ \text{kg}$ Initial speed, $u = 5000\ \text{m/s}$ Final speed, $v = 0$ Time, $t = 2\ \text{min} = 120\ \text{s}$

Formula: $F_{av} = \frac{m(v-u)}{t}$

Substitution: $F_{av} = \frac{4\times10^5\,(0-5000)}{120}$ $F_{av} = -\frac{2\times10^9}{120} = -1.67\times10^7\ \text{N}$

Result: The average force applied by the ground on the aeroplane is $1.67\times10^7\ \text{N}$, directed opposite to the motion of the jet.

The aeroplane's momentum is not lost; it is transferred to the Earth. The ground exerts a retarding force on the aeroplane and, by Newton's third law, the aeroplane exerts an equal and opposite force on the ground for the same time. So the Earth gains exactly the momentum the aeroplane loses. The total momentum of the aeroplane–Earth system is unchanged, so the law of conservation of momentum holds. The Earth's mass is so large that its resulting velocity is negligible.

(a) From the velocity–time graph, the box accelerates uniformly from rest to $5\ \text{m s}^{-1}$ in the first $2.0\ \text{s}$ (the slope of this straight section is the initial acceleration):

$a=\frac{v-u}{t}=\frac{5-0}{2.0}=2.5\ \text{m s}^{-2}$

For the unbalanced force $F=10\ \text{N}$, Newton's second law gives

$m=\frac{F}{a}=\frac{10}{2.5}=4\ \text{kg}$

After $2.0\ \text{s}$ the velocity–time graph is horizontal, so the box moves with constant velocity and its acceleration is zero. The resultant force is therefore zero:

$F-f=ma=0\ \Rightarrow\ f=F=12\ \text{N}$

The frictional force is $12\ \text{N}$, acting opposite to the motion.

The distance travelled equals the area under the velocity–time graph from $0$ to $8.0\ \text{s}$. The box accelerates from $0$ to $5\ \text{m s}^{-1}$ in the first $2.0\ \text{s}$ (a triangle), then moves at constant $5\ \text{m s}^{-1}$ (a rectangle):

$s=\tfrac12\times 2.0\times 5+(8.0-2.0)\times 5=5+30=35\ \text{m}$

The box travels $35\ \text{m}$ in $8.0\ \text{s}$.

No. By Newton's first law, zero net force means zero acceleration, not zero velocity. If the body is already moving, it continues to move with constant velocity in a straight line. The body is at rest only if it was at rest to begin with. For example, a hockey puck sliding on smooth ice, or a car cruising at constant speed on a straight road, has zero net force but is not at rest.

Let $m_1=7\ \text{kg}$ and $m_2=12\ \text{kg}$, $g=9.8\ \text{m s}^{-2}$. The heavier mass moves down and the lighter one up, with the same acceleration $a$ and the same tension $T$.

For $m_2$ (down): $m_2g-T=m_2a$ For $m_1$ (up): $T-m_1g=m_1a$

Adding,

$a=\frac{(m_2-m_1)g}{m_1+m_2}=\frac{(12-7)\times9.8}{19}\approx 2.58\ \text{m s}^{-2}$

Tension:

$T=\frac{2m_1m_2\,g}{m_1+m_2}=\frac{2\times7\times12\times9.8}{19}\approx 86.7\ \text{N}$

Check: $T=m_1(g+a)=7\times(9.8+2.58)=86.7\ \text{N}$.