- #1
songoku
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- Homework Statement
- A rocket is launched upwards with initial speed of 120 m/s. After moving for 0.5 km, an object is released from the rocket. Given the rocket does not experience change in velocity, what is the distance between them after 10 seconds?
- Relevant Equations
- Conservation of momentum
Kinematics
Speed of rocket after moving 0.5 km = ##\sqrt{u^2-2gh}=\sqrt{120^2 - 2 \times 9.81 \times 500}=3\sqrt{510}## m/s
Then I try to consider conservation of momentum to find the speed of the object after being released.
Total momentum before the object is released = total momentum after the object released
Let:
m1 = mass of rocket
m2 = mass of object
v1 = velocity of rocket after object being released = u (initial velocity) = ##3\sqrt{510}## m/s
v2 = velocity of object after being released
So,
$$m_1.u + m_2.u=m_1.v_1+m_2.v_2$$
$$m_2.u=m_2.v_2$$
$$u=v_2$$
For momentum to be conserved, it means that after released the object will move upwards with the same speed as the rocket so the distance between them after 10 seconds will be zero?
Thanks
Then I try to consider conservation of momentum to find the speed of the object after being released.
Total momentum before the object is released = total momentum after the object released
Let:
m1 = mass of rocket
m2 = mass of object
v1 = velocity of rocket after object being released = u (initial velocity) = ##3\sqrt{510}## m/s
v2 = velocity of object after being released
So,
$$m_1.u + m_2.u=m_1.v_1+m_2.v_2$$
$$m_2.u=m_2.v_2$$
$$u=v_2$$
For momentum to be conserved, it means that after released the object will move upwards with the same speed as the rocket so the distance between them after 10 seconds will be zero?
Thanks