An object of mass 3.0 kg travelling at a speed of 6.0 m/s, collides with an object of mass 2.0 kg, travelling in the opposite direction at a speed of 2.0 m/s. The objects stick together during the collision. What is the speed and direction of the combined mass after the collision?
A) 4.4 m/s to the left
B) 4.4 m/s to the right
Answers
Answer :
Velocity of combined mass after the collision is 2.8 m/s to the right .
Step by step Explanation :
We have ,
- Velocity of
- Velocity of
Let the velocity of combined be V.
As no external force acts on the system
So According to the law of conservation of Momentum .
Intital momentum = Final momentum
Put the given values, then
______________
More About the topic :
• Conservation of Momentum :
If no external force acts on the system of constant mass the total momentum of the system remains constant .
Answer:
Answer :
Velocity of combined mass after the collision is 2.8 m/s to the right .
Step by step Explanation :
We have ,
\sf\:m_1=3kgm
1
=3kg
\sf\:m_2=2kgm
2
=2kg
Velocity of \sf\:m_1,\:v_1=6ms^{-1}\hat{i}m
1
,v
1
=6ms
−1
i
^
Velocity of \sf\:m_2,\:v_2=2ms^{-1}\hat{(-i)}m
2
,v
2
=2ms
−1
(−i)
^
Let the velocity of combined be V.
As no external force acts on the system
So According to the law of conservation of Momentum .
Intital momentum = Final momentum
\sf\:m_1v_1+m_2v_2=(m+M)Vm
1
v
1
+m
2
v
2
=(m+M)V
Put the given values, then
\sf\implies\:3\times6+2(-2)=(3+2)V⟹3×6+2(−2)=(3+2)V
\sf\implies\:18-4=5V⟹18−4=5V
\sf\implies\:14=5V⟹14=5V
\sf\implies\:V=\dfrac{14}{5}⟹V=
5
14
\sf\implies\:V=2.8ms^{-1}⟹V=2.8ms
−1
______________
More About the topic :
• Conservation of Momentum :
If no external force acts on the system of constant mass the total momentum of the system remains constant .