Question

A rocket of mass 5700 kg ejects mass at a constant rate of 15 kg/s with constant speed of 12 km/s . What will be the acceleration of the rocket 1 minute ater the blast ?

Answer 1

Force on the rocket:
F= v dm/dt
= 12000 x 15= 180000 N

Mass after 1 min
M= m - dm/dt x t
= 5700 - 15 x 60
= 4800 kg

Accelration
a= F/m
or a= F-mg/M
= 180000 - (4800 x 10)/ 4800
=27.5 m/s2

Answer 2

The acceleration of the rocket is $27.5 \ m/s^2$

Given:

Mass of rocket = 5700 kg  

Rate of flow = dm/dt = 15 kg/s  

Speed of flow = v = 12 km/s = $12 \times 10^{3} \ m/s$

Solution:  

The force of ejection is given by the formula given below:

$F=v \frac{d m}{d t}$

$F=\left(12 \times 10^{3}\right)(15)$

$\Rightarrow F=180000 N$

Thereby,

Mass of ejection after 1 minute = Total ejection mass – Mass of ejection during 1 minute

$M=m-\left(\frac{d m}{d t}\right) t$

$M=5700-(15 \times 60)$

We know that,  

1 minute = 60 seconds

$\Rightarrow M=5700-900$

$\therefore M=4800 \ \mathrm{kg}$

Therefore, acceleration of the ejection is given by the formula below:

$a=\frac{F-M g}{M}$

$a=\frac{180000-(4800 \times 9.8)}{4800}$

$\therefore a=27.7 \ \mathrm{m} / \mathrm{s}^{2}$