• Physics 9702 Doubts | Help Page 10

Question 55: [Momentum > Projectile motion]

A projectile of mass m is fired with velocity v from a point P, as shown below. 

Neglecting air resistance, the magnitude of the change in momentum between leaving P and arriving at Q is

A zero             B ½ mv           C mv√2           D mv               E 2mv

Reference: Past Exam Paper – N80 / II / 6

Solution 55:

Velocity, v is a vector has can be resolved into 2 perpendicular components as follows:

Define the upwards direction as positive for the vertical and the direction from P to Q as positive for the horizontal.

At P,

Horizontal component of velocity = (+) v cos45^o

Vertical component of velocity = (+) v sin45^o  

Similarly, momentum, p is also a vector and can be resolved into 2 perpendicular components.

At P,

Horizontal component of momentum = (+) mv cos45^o

Vertical component of momentum = (+) mv sin45^o  

Resultant momentum = [(mv cos45^o)² + (mv sin45^o)²]^0.5 = mv

Now, arriving at Q, velocity is NOT zero (it may be zero at the instant just after hitting the ground but not when arriving at Q).

Since air resistance is negligible, the horizontal component of velocity at Q is the same as at P. However, the vertical component of the velocity is now downwards due to gravity (its magnitude is the same as at P since the (maximum) vertical distance travelled and the acceleration (due to gravity) are the same).

At Q,

Horizontal component of velocity = (+) v cos45^o

Vertical component of velocity = (-) v sin45^o  

Horizontal component of momentum = (+) mv cos45^o

Vertical component of momentum = (-) mv sin45^o  

Change in (resultant) momentum = mΔv = m (v_f – v_i)

where v_f is the final velocity and v_i is the initial velocity

The horizontal velocity (and hence, horizontal momentum) is unchanged (change = 0).

Change in vertical component of momentum = (-)mv sin45^o – mv sin45^o  = - 2mv sin45^o  

Change in (resultant) momentum = [0² + (- 2mv sin45^o)²]^0.5 = [4 x (mv)² x 0.5]^0.5

Change in (resultant) momentum = [2 x (mv)²]^0.5 = mv √2

Answer: C

Question 56: [Kinematics > Graphs]

The graph shows the variation with time t of the velocity v of a bouncing ball, released from rest. Downward velocities are taken as positive.

At which time does the ball reach its maximum height after bouncing?

Reference: Past Exam Paper – J99 / I / 3

Solution 56:

The graph is that of a velocity-time graph. The area enclosed under the graph (= v x t) gives the displacement.

The ball is released from rest (from a height) and downward velocities are taken as positive. So, the upper area in the graph shows the distance fallen (not height) while the area enclosed by the graph under the x-axis gives the height.

From the graph, it can be seen that speed increases from t = 0 to t = A (the ball falls under the acceleration due to gravity).

From t = A to t = B, there is a steep decrease in speed and from t = B to t = C, the speed increases in the opposite direction (upward). So, t = A to t = C represents the time of collision of the ball with the ground.

At t = C, the (upward) speed is maximum and it decreases (due to the acceleration due to gravity acting downwards) until t = D (speed is zero, so maximum height has been reached).

Answer: D

Question 57: [Momentum]

A group of students investigating the principle of conservation of momentum use a small truck travelling over a frictionless surface.

Sand is dropped into the truck as it passes X. At Y, a trapdoor in the bottom of the truck opens and the sand falls out.

How does the velocity of the truck change when the sand is added to the truck at X and then leaves the truck at Y?

Reference: Past Exam Paper – November 2011 Paper 12 Q10

Solution 57:

Answer: B.

In this problem, you should consider conservation of the total momentum of the truck and sand, rather than considering the truck in isolation. 

From conservation of momentum (p = mv), when mass (sand) is added to the truck, the total mass (of the truck and sand) increases, resulting in speed v (of the moving truck containing the sand) to decrease so that momentum p is unchanged (conserved).  So, at X, the speed of the truck decreases.

Then the truck (containing the sand) will move with this new speed.

Consider at Y where the sand falls. So, the mass of truck has returned to normal since it no longer contains the sand. But, as the sand falls [it’s obviously falling at some speed which is equal to the speed of the truck], it has some momentum such that the sum of momentum of the (separate) truck and sand together is unchanged and equal to the total momentum at X. So, at Y, the speed of the truck remains the same.

So, what is important to note from this question is that at X, the sand is in the truck which is moving. So, they can be considered as a single body. But at Y, the sand falls from the truck. Now, they are 2 separate bodies. But since the sand was previously in the truck which was moving, it falls at the same speed of motion of the truck.  So, the momentum of the sand should also be considered, in addition to the momentum of the truck.

Question 58: [Projectile motion]

When a rifle is fired horizontally at a target P on a screen at a range of 25m, the bullet strikes the screen at a point 5.0mm below P. The screen is now moved to a distance of 50m and the rifle again fire horizontally at P in its new position.

Assuming that air resistance may be neglected, what is the new distance below P at which the screen would now be struck?

A 5√2mm        B 10mm          C 15mm          D 20mm          E 25mm

Reference: Past Exam Paper – November 1982 / II / 4

Solution 58:

Answer: D.

The equations for uniformly accelerated motion involving displacement are

s = ut + ½at²    and      v² = u² +2as

We need to find a vertical displacement. For the first equation, t is not known. For the second equation, v is not known. So, the calculation cannot be carried out directly. Since there are 2 cases, we can write separate equations and solve them simultaneously.

For case 1,

Let (horizontal) velocity with which the bullet leaves the rifle = v

Let time taken for bullet to travel the horizontal distance of 25m = t₁

Speed = distance / time

Time, t₁ = 25 / v

Now, consider the vertical motion of the bullet,

Initial (vertical) velocity of bullet, u = 0

Acceleration, a = g, the acceleration due to gravity

Displacement, s = 5.0mm = 5.0x10^-3m

Time to be displaced by vertical displacement s, t₁ = 25 / v

s = ut + ½ t²

5.0x10^-3 = 0 + 0.5 g (25 / v)²  ……………….(1)

For case 2,

Time to travel (horizontal) distance of 50m, t₂ = 50 / v

Let the new displacement below P = x

s = ut + ½at²

 x = 0 + 0.5 g (50 / v)² ………………………(2)

To find x, divide equation (2) by equation (1)

x / (5x10^-3) = (50 / 25)² = 4

New displacement, x = 20.0x10^-3 = 20mm

Question 59: [Projectile motion]

A ball, dropped on a 45^o inclined plane, makes an elastic collision with the surface. Stroboscopic photographs (a series of exposures on the same film at equal time intervals) are taken of the path of the ball. Which one of the following diagrams best represents the photographs?

Reference: Past Exam Paper – June 1984 / II / 3

Solution 59:

Answer: A.

Under the influence of acceleration due to gravity, the (downward) speed of the ball increases every second. So, in a specific time interval, the vertical separation of the exposures should increase (since the distance travelled every second increases).

Just after the collision, the vertical speed is zero and the horizontal speed is equal to the speed just before collision [for elastic collision, relative speed of approach is equal to the relative speed of separation]. So, the separation of exposures just after collision should be equal to that just before collision. [B is eliminated]

The horizontal speed is not affected by any forces (like air resistance), the horizontal separation should be equal. [C and E are eliminated]

Let’s assume that each exposure was taken after 1 second each. Since there are 8 exposures, the final speed before collision would be v = u + at = 0 + g8 = 8g. After the collision, the horizontal sped is constant and equal to 8g (as explained before). However, the vertical speed of collision is zero and will only be equal to the horizontal speed at the 8^th exposure. In such a situation, if a line is drawn between 2 exposures there, it should be parallel to the inclined plane [since horizontal and vertical components are equal]. It also means that for the earlier exposures, the horizontal component is more dominant. 

But looking at diagram A and D, it is seen that for D, a parallel line between 2 exposures is obtained before the 8^th exposure is reached. [So, D is also incorrect]. And if we look at diagram A, it is seen clearly that the horizontal is more dominant than the vertical for the earlier exposures.