A hot-air balloon is moving vertically upwards with a constant speed of 3.00 m s-1.

Question 37

A hot-air balloon is moving vertically upwards with a constant speed of 3.00 m s^-1. A sandbag is dropped from the balloon. It takes 5.00 s for the sandbag to fall to the ground.

 

What was the height of the balloon when the sandbag was released?

A 29 m                        B 108 m                      C 123 m                      D 138 m

 

 

 

 

 

Reference: Past Exam Paper – November 2017 Paper 12 Q6

 

 

 

 

 

Solution:

Answer: B.

The balloon is moving upwards while the sandbag is falling downwards. We have motion in both the upward and downward directions.

 

Let the upward motion be positive.

 

Initial speed of sandbag: u = 3.00 m s^-1

Acceleration due to gravity: a = - 9.81 m s^-2 (downwards)

Time taken by sandbag to fall: t = 5.00 s

 

Equation of motion: s = ut + ½ at²

s = (3×5) + ½ × -9.81 × 5² = (-) 107.7 = (-) 108 m

A metal ball of mass 40 g falls vertically onto a spring, as shown in Fig. 4.1.

Question 17

A metal ball of mass 40 g falls vertically onto a spring, as shown in Fig. 4.1.

Fig. 4.1 (not to scale)

 

The spring is supported and stands vertically. The ball has a speed of 2.8 m s^-1 as it makes contact with the spring. The ball is brought to rest as the spring is compressed.

 

(a) Show that the kinetic energy of the ball as it makes contact with the spring is 0.16 J. [2]

 

 

(b) The variation of the force F acting on the spring with the compression x of the spring is shown in Fig. 4.2.

Fig. 4.2

 

The ball produces a maximum compression XB when it comes to rest. The spring has a spring constant of 800 N m^-1.

 

Use Fig. 4.2 to

(i) calculate the compression XB, [2]

 

(ii) show that not all the kinetic energy in (a) is converted into elastic potential energy in the spring. [2]

 

 

 

 

 

Reference: Past Exam Paper – June 2015 Paper 21 Q4

 

 

 

 

 

Solution:

(a)

kinetic energy = ½ mv²            

kinetic energy = ½ × 0.040 × (2.8)² = 0.157 J or 0.16 J

 

 

(b)

(i)

{From Hooke’s law,}

k = F / x                       or F = kx        

 

{x = F / k}

XB = 14 / 800

XB = 0.0175 m                                    

 

(ii)

{The elastic potential energy stored in the spring can be obtained from the area under the F-x graph.}

 

area under graph = elastic potential energy stored

or Elastic PE = ½ kx²                 or ½ Fx

 

{Elastic PE = ½ Fx = ½ × 14 × 0.0175}

(energy stored =) 0.1225 J less than KE (of 0.16 J)

 

{This value of elastic PE is less than the KE calculated above.}