A body of mass 2.0kg makes an elastic collision with another body at rest and continues to move in the original direction but with 1/4 of its original speed.a) What is the mass of the other body?b) What is the speed of the two-body center of mass if the initial speed of the 2.0kg body was 4.0m/s?

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Asked Nov 6, 2019
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A body of mass 2.0kg makes an elastic collision with another body at rest and continues to move in the original direction but with 1/4 of its original speed.

a) What is the mass of the other body?

b) What is the speed of the two-body center of mass if the initial speed of the 2.0kg body was 4.0m/s?

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Expert Answer

Step 1

Both Momentum and Kinetic energy will be conserved in this system. Thus, from the given conditions:

From Momentum, we have:
Initial Momentum = Final Momentum
ти + ти,
— тv, + тv,
or, 2u,0 2-m,v^
и,
or, 2u
[Its given, v
m2v2
2
Зи,
or, mv2
2
-- Eq
Eql
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From Momentum, we have: Initial Momentum = Final Momentum ти + ти, — тv, + тv, or, 2u,0 2-m,v^ и, or, 2u [Its given, v m2v2 2 Зи, or, mv2 2 -- Eq Eql

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Step 2

a)Thus, the mass of the other body will be:

From Kinetic Energy, we have:
Initial Kinetic Energy = Final Kinetic Energy
1
1
2
1
1
2
2
-ти* +
2
-туf +- т,v,
2
-ти, 3 —
2
Зи,
и
2
or, 2u,0 2
16
2
Putting the value of Eql i.e m2v^ =(m,v2)v,]
5u
---- Eq2
or,
Now, putting the value of v, in Eqi, we get:
5u, _Зи,
т,
4
2
or, m = 1.2 kg
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From Kinetic Energy, we have: Initial Kinetic Energy = Final Kinetic Energy 1 1 2 1 1 2 2 -ти* + 2 -туf +- т,v, 2 -ти, 3 — 2 Зи, и 2 or, 2u,0 2 16 2 Putting the value of Eql i.e m2v^ =(m,v2)v,] 5u ---- Eq2 or, Now, putting the value of v, in Eqi, we get: 5u, _Зи, т, 4 2 or, m = 1.2 kg

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Step 3

(b) The speed of the two-b...

The speed of the center of mass is given by:
ти, + ти,
VCM
2x4+0
2 1.2
т + т,
or, Va 2.5m /s
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The speed of the center of mass is given by: ти, + ти, VCM 2x4+0 2 1.2 т + т, or, Va 2.5m /s

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