Saturday, November 8, 2014

Physics 9702 Doubts | Help Page 9

  • 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) = 8.75x105Ns

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.
8.75x105 = 1200 (–20) + 2300 (v)
2300v = (8.75x105) + 24000
New speed of rocket, v = (8.99x105) / 2300 = 390.87 = 390ms-1






Question 51: [Forces > Vectors]
Diagram shows barrel of weight 1.0 × 103N on frictionless slope inclined at 30o 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 × 103J                B 4.3 × 103J                C 5.0 × 103J                D 1.0 × 104J

Reference: Past Exam Paper – November 2013 Paper 11 & 12 Q17



Solution 51:
17 – Ans: A.


Component of weight along (down) the slope = (1.0x103)sin30o.

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 = (1.0x103)sin(30) x 5 = 2.5 x 103 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

(c) Two parallel strings S1 and S2 are attached to disc of diameter 12 cm, as shown.
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 S1 and S2. 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 S1

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 S1 and S2 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 S1, 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 60o to the horizontal.}
The horizontal speed is constant at 8.2ms-1 {since air resistance is negligible}
The vertical component of speed = 8.2 tan60o = 14.2ms-1 

(ii)
{Consider the vertical motion. v2 = u2 + 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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