a body of mass 30 m per second collides another body mass m moving in the same direction with a velocity of 20 m per second if both body is now moving in same direction with a velocity of 10 m per second calculate the value of m
Answers
Answer:
search-icon-header
Search for questions & chapters
search-icon-image
Class 11
>>Physics
>>Work, Energy and Power
>>Collisions in Two Dimensions
>>A body of mass 2 kg moving ...
Question
Bookmark
A body of mass 2 kg moving with a velocity 3 m/s collides with a body of mass 1 kg moving with a velocity of 4 m/s in opposite direction. If the collision is head on an completely inelastic, then :
This question has multiple correct options
Medium
Solution
verified
Verified by Toppr
Correct option is
A
both particles move together with velocity (2/3) m/s
B
the momentum of system is 2 kg m/s throughtout
D
the loss of KE of system is (49/3) J
Given : m
1
=2kg m
2
=1kg u
1
=3 m/s u
2
=4 m/s
Let the common velocity of the combined body be V
Mass of combined body M=2+1=3kg
Applying conservation of momentum :
2×3−1×4=3×V ⟹V=
3
2
m/s
Momentum of the system P=2×3−1×4=2 kg m/s
Loss in kinetic energy of the system due to collision ΔK.E=K.E
i
−K.E
f
ΔK.E=
2
1
(2)×3
2
+
2
1
(1)×4
2
−
2
1
(3)×(
3
2
)
2
⟹ΔK.E=
3
49
J
Explanation:
please brainliest answer
Given : m
1
=2kg m
2
=1kg u
1
=3 m/s u
2
=4 m/s
Let the common velocity of the combined body be V
Mass of combined body M=2+1=3kg
Applying conservation of momentum :
2×3−1×4=3×V ⟹V=
3
2
m/s
Momentum of the system P=2×3−1×4=2 kg m/s
Loss in kinetic energy of the system due to collision ΔK.E=K.E
i
−K.E
f
ΔK.E=
2
1
(2)×3
2
+
2
1
(1)×4
2
−
2
1
(3)×(
3
2
)
2
⟹ΔK.E=
3
49
J
