Physics · Work, Power & Energy
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When the momentum of body is increased by 200%, its kinetic energy increases by:
- A
200 %
- B
300 %
- C
400 %
- D
800 %
Momentum P = mv
Kinetic energy K.E= 1 /2 mv^2
K.E= p^2/ 2m
If p is increased vy 200% ,then new p will be.
p*= p+ 200%p
p* = p+2p
p* = 3p
In this case, new K.E will be,
K.E* = p*^2 / 2m
= [3p]^2 /2m
= 9p^2/ 2m
K.E*= 9 K.E [as, p^2/ 2m= K.E]
ΔK.E% [ percent change in K.E] = K.E* -K.E
= 9 K.E - K.E
= 8 *100% ΔK.E% = 800%
This option is incorrect.
Momentum P = mv
Kinetic energy K.E= 1 /2 mv^2
K.E= p^2/ 2m
If p is increased vy 200% ,then new p will be.
p*= p+ 200%p
p* = p+2p
p* = 3p
In this case, new K.E will be,
K.E* = p*^2 / 2m
= [3p]^2 /2m
= 9p^2/ 2m
K.E*= 9 K.E [as, p^2/ 2m= K.E]
ΔK.E% [ percent change in K.E] = K.E* -K.E
= 9 K.E - K.E
= 8 *100% ΔK.E% = 800%
This option is incorrect.
Momentum P = mv
Kinetic energy K.E= 1 /2 mv^2
K.E= p^2/ 2m
If p is increased vy 200% ,then new p will be.
p*= p+ 200%p
p* = p+2p
p* = 3p
In this case, new K.E will be,
K.E* = p*^2 / 2m
= [3p]^2 /2m
= 9p^2/ 2m
K.E*= 9 K.E [as, p^2/ 2m= K.E]
ΔK.E% [ percent change in K.E] = K.E* -K.E
= 9 K.E - K.E
= 8 *100% ΔK.E% = 800%
This option is incorrect.
Momentum P = mv
Kinetic energy K.E= 1 /2 mv^2
K.E= p^2/ 2m
If p is increased vy 200% ,then new p will be.
p*= p+ 200%p
p* = p+2p
p* = 3p
In this case, new K.E will be,
K.E* = p*^2 / 2m
= [3p]^2 /2m
= 9p^2/ 2m
K.E*= 9 K.E [as, p^2/ 2m= K.E]
ΔK.E% [ percent change in K.E] = K.E* -K.E
= 9 K.E - K.E
= 8 *100% ΔK.E% = 800%
Momentum P = mv
Kinetic energy K.E= 1 /2 mv^2
K.E= p^2/ 2m
If p is increased vy 200% ,then new p will be.
p*= p+ 200%p
p* = p+2p
p* = 3p
In this case, new K.E will be,
K.E* = p*^2 / 2m
= [3p]^2 /2m
= 9p^2/ 2m
K.E*= 9 K.E [as, p^2/ 2m= K.E]
ΔK.E% [ percent change in K.E] = K.E* -K.E
= 9 K.E - K.E
= 8 *100% ΔK.E% = 800%
Tagged under Physics · Work, Power & Energy · 2021