A block of mass m=3.6 kg, moving on africtionless surface with a speed v_1=9.3 m/s makes a perfectly elastic collision with a block of mass Mat rest.

sibuzwaW

sibuzwaW

Answered question

2021-02-23


A block of mass m=3.6 kg, moving on africtionless surface with a speed v1=9.3 m/s makes a perfectly elastic collision with a block of mass Mat rest. After the collision, the 3.6 kg block recoils with a speed of v1=2.7 m/s in figure, the speed of the vlock of mass M after the collision is closest to:
a. 9.3 m/s
b. 6.6 m/s
c. 8.0 m/s
d. 10.7 m/s
e. 12.0 m/s

Answer & Explanation

Margot Mill

Margot Mill

Skilled2021-02-25Added 106 answers

Let:
m=3.6kg
vi=9.3ms
vf=2.7ms
V=to be found
M=to get rid of
Conservation of momentun:
p initial=p final:
mvi=mvf+MV
Solving for M:
MV=mvimvf
M=mvimvfV=m(vivf)V
Conservation of kinetic energy:
KE initial=KE final:
.5vi2=.5mvf2+.5MV2
Getting rid of the .5
mvi2=mvf2+MV2
Solving for M:
MV2=mvi2mvf2
M=mvi2mvf2V=m(vi2vf2}{V}
Setting both expressions for M equal:
m(vivf)V=m(vi2vf2)V2
Canceling out m and knowing that (vi2vf2) is the difference of two squares:
vivfV=(vivf)(vi+vf)V2
Solving for V:
V=vi+vf=(9.3ms)+(2.7ms)=6.6ms

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