# Physics 9702 Doubts | Help Page 9

__Question 50: [Momentum > Rocket]__
A rocket of total mass 3500kg is
moving at 250ms

^{-1}through space. When the booster rockets are fired 1200kg of burnt fuel is ejected from the back of the rocket at 20ms^{-1}. What is the new speed of the rocket?**Reference:**

*Physics For You - End-of-Chapter Questions*

__Solution 50:__
Define the forward direction as the
positive direction.

__Before firing of booster rockets__,

(Momentum = mass x velocity)

Sum of momentum before firing = 3500
(250) = 8.75x10

^{5}Ns__After firing of booster rockets__,

Since the 1200kg of burnt fuel is
fired from the back of the rocket, its direction of motion is backward. So, its
velocity = – 20ms

^{-1}.
New mass of rocket = 3500 – 1200 =
2300kg

Let new speed of rocket = v

From the

**conservation of momentum**, sum of momentum before firing of the booster rockets should be equal to the sum of momentum after firing of the booster rockets.
8.75x10

^{5}= 1200 (–20) + 2300 (v)
2300v = (8.75x10

^{5}) + 24000
New speed of rocket, v = (8.99x10

^{5}) / 2300 = 390.87 =**390ms**^{-1}

__Question 51: [Forces > Vectors]__
Diagram shows barrel of weight 1.0 ×
10

^{3}N on frictionless slope inclined at 30^{o}to horizontal.
Force is applied to barrel to move
it up slope at constant speed. Force is parallel to slope. What is work done in
moving barrel a distance of 5.0 m up slope?

A 2.5 × 10

^{3}J B 4.3 × 10^{3}J C 5.0 × 10^{3}J D 1.0 × 10^{4}J**Reference:**

*Past Exam Paper – November 2013 Paper 11 & 12 Q17*

__Solution 51:__
17 – Ans: A.

Component of weight along (down) the
slope = (1.0x10

^{3})sin30^{o}.
A force equal in magnitude (but [up]
along the slope – as indicated by ‘force’ in the diagram) is required to keep
the barrel stationary.

Since the barrel is to be moved up
the slope at

__constant speed__, the**resultant acceleration**up the slope should be**zero**. Thus, the resultant force is zero and so, the ‘force’ up the slope should be equal to the component of the weight along (down) the slope.
Work done = Force x distance moved
(in direction of force)

Work done = (1.0x10

^{3})sin(30) x 5 = 2.5 x 10^{3}J

__Question 52: [Momentum > Cannon]__
A cannon fire a cannonball of mass
55kg at 35ms

^{-1}. The cannon recoils at 2.5ms^{-1}.**(a)**What is the mass of the cannon?

**(b)**If the cannonball becomes embedded in a target of mass 600kg, at what speed does the target move immediately after the impact?

**Reference:**

*Physics For You - End-of-Chapter Questions*

__Solution 52:__**(a)**

From the

**conservation of momentum**, the momentum of the cannonball should be equal to the momentum of the cannon at the__instant__the cannonball is fired. [At the instant the cannonball is fired, the cannon and the cannonball can be considered to be 2 different objects, unlike the alternative method to solve this problem discussed next.]
(Momentum = mass x velocity)

Momentum of cannonball = 55 (35) =
1925Ns

Let mass of cannon = M

Momentum of cannon = 2.5M

2.5M = 1925

Mass of cannon, M = 1925 / 2.5 =

**770kg**
[An alternative method to solve this problem would be to
consider the sum of momentum before and after the firing of the cannon.

Before the firing of the cannonball, the mass of the cannon
and that of the cannonball can be considered as one object. But since they are
initially stationary (not moving), their momentum before the firing is zero.

After the firing, the momenta of the cannon and the cannon
ball must be considered separately. Since the cannon recoils (gets jerk back)
when the cannonball is fired forward, one of the speed must be taken to be
negative, depending on how the positive direction has been defined.

The mass of the cannon can then be obtained by equation the
sum of momentum before the firing to the sum of momentum after the firing.]

**(b)**

Before impact, momentum of
cannonball = 55 (35) = 1925Ns

The cannonball becomes embedded in a
target of mass 600kg. So, after the impact, both of them move as a single body.

Mass of target + cannonball = 600 +
55 = 655kg

Let speed of the target + cannonball
= v

Assuming the target was initially at
rest, its speed is zero and its momentum is also zero.

From

**conversation of momentum**, sum of momentum before impact should be equal to the sum of momentum after the impact.
1925 = 655 (v)

Speed of the target + cannonball, v
= 1925 / 655 = 2.9389 =

**2.94ms**^{-1}

__Question 53: [Force > Moment]__**(a)**Define moment of force

**(b)**State two conditions necessary for body to be in equilibrium

Disc is free to rotate about axis
normal to its plane. Axis passes through centre C of disc. Lever of length 30cm
is attached to disc. When force F is applied at right angles to lever at its
end, equal forces are produced in S

_{1}and S_{2}. Disc remains in equilibrium.
(i) On Fig, show direction of force
in each string that acts on disc

(ii) For force F of magnitude 150 N,
determine

1. Moment of force F about centre of
disc

2. Torque of couple produced by
forces in strings

3. Force in S

_{1}**Reference:**

*Past Exam Paper – November 2003 Paper 2 Q3*

__Solution 53:__**(a)**

The moment of a force is defined as
the product of the force and the perpendicular distance (of the force) from the
pivot

**(b)**

There should be no resultant force (in
any direction) {not only upward and downward forces}

There should be no resultant moment
(about any point)

**(c)**

(i)

The correct direction in both should
be shown {as in the diagram below}

{Force F would cause an
anticlockwise moment. To keep the disc in equilibrium, the tensions in S

_{1}and S_{2}should be as indicated below, such that they produce a clockwise moment to counters the moment due to force F.
Also, tensions (which are
the reaction forces – as stated by Newton’s third law) in strings are usually
away from the object, with a magnitude equal to the force present, but in an
opposes the latter}

(ii)

1. Moment of force F (= F x distance)
= 150 x 0.3 = 45Nm

2. {For
equilibrium, this should be equal to the moment produces by force F.}

Torque of the couple = 45Nm

3. {Torque =
(Magnitude of 1 Force) x distance between the 2 forces producing the torque}

45 = 0.12 x T

Force in S

_{1}, T (= 45 / 0.12) = 375N

__Question 54: [Projectile motion]__
Ball is thrown horizontally from top
of building, as shown.

Ball is thrown with horizontal speed
of 8.2 m s

^{–1}. Side of building is vertical. At point P on path of ball, ball is distance x from building and is moving at an angle of 60° to horizontal. Air resistance is negligible.**(a)**For ball at point P,

(i) Show that vertical component of
its velocity is 14.2 m s

^{–1}
(ii) Determine vertical distance
through which ball has fallen

(iii) Determine horizontal distance
x

**(b)**Path of ball in (a), with initial horizontal speed of 8.2 m s

^{–1}, is shown again. On Fig, sketch new path of ball for ball having initial horizontal speed

(i) Greater than 8.2 m s

^{–1}and with negligible air resistance (label path G)
(ii) Equal to 8.2 m s

^{–1}but with air resistance (label path A)**Reference:**

*Past Exam Paper – November 2010 Paper 21 Q2*

__Solution 54:__**(a)**

(i)

{At point P, a triangle
may be drawn for the components of the speed, with the horizontal (component
of) speed = 8.2ms

^{-1}, the vertical needs to be found and the hypotenuse being at an angle of 60^{o}to the horizontal.}
The horizontal speed is constant at
8.2ms

^{-1}{since air resistance is negligible}
The vertical component of speed =
8.2 tan60

^{o}= 14.2ms^{-1}
(ii)

{Consider the vertical
motion. v

^{2}= u^{2}+ 2as = 0 + 2as}
14.22 = 0 + 2 (9.8) h

Vertical distance fallen = 10.3m

(iii)

{First, find the time, t
at which the ball has fallen to vertical distance calculated previously. For
the vertical motion, v = u + at = 0 + at}

Time of descent of ball, t (= v / a)
= 14.2 / 9.8 = 1.45s

{Now, for this amount of
time, the horizontal distance travelled by the ball moving at (horizontal) speed
= 8.2ms

^{-1}. [Speed = Distance / Time] Distance = Speed x Time}
Horizontal distance x = 8.2 x 1.45 =
11.9m

**(b)**

(i)

The sketch shows a smooth path
curved and above the given path. The ball hits the ground at a more acute angle

{The ball would travel a
greater horizontal distance at any instant of time as the horizontal speed is
greater. Note that vertical component of velocity behaves in the same way as
previously. The horizontal component is greater than before (and constant at
all time), so the resultant velocity will be at an angle more acute than before.}

(ii)

The sketch shows a smooth path
curved and below the given path. The ball hits the ground at a steeper angle.

{Air resistance affects
all components of the speed, reducing them, since it opposes motion. So, a
smaller horizontal distance is travelled by the ball.

The vertical component of
speed undergoes acceleration (due to gravity) until terminal speed is reached (ball
may hit the ground before terminal velocity is reached) but compare to the
cases before, the speed increases at a smaller rate due to air resistance.

As for the horizontal
component, it decreases due to air resistance. So, the resultant speed will be
at a steeper angle to the ground since the horizontal component is smaller than
in the previous cases.}

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