a stone is projected vertically up to reach maximum height h. the stone passes points P and Q
Answers
Explanation:
\bold{\underline{\underline{\huge{\rm{AnsWer:}}}}}
AnsWer:
Kinetic energy of mass 6 kg is 192 Joules.
\bold{\underline{\underline{\large{\rm{StEp\:by\:stEp\:explanation:}}}}}
StEpbystEpexplanation:
GiVeN :
A bomb of mass 9 kg.
The bomb explodes into two pieces
\bold{m_1}m
1
= 3 kg.
\bold{m_2}m
2
= 6 kg.
Velocity of \bold{m_1}m
1
= 16 m/sec.
Initial Velocity (u) = 0 m/sec
\bold{u_1}u
1
= 0 m/sec
\bold{u_2}u
2
= 0 m/sec
To FiNd :
Kinetic energy of mass 6 kg i.e \bold{m_2}m
2
SoLuTiOn :
We need to first calculate the velocity of \bold{m_2}m
2
using conservation of linear momentum.
\bold{\sf{Total\:initial\:momentum\:=\:Total\:Final\:Momentum}}Totalinitialmomentum=TotalFinalMomentum
\rightarrow→ \bold{\sf{m_1\:\times\:u_1\:+\:m_2\:\times\:u_2\:=\:m_1\:\times\:v_1\:+\:m_2\:\times\:v_2}}m
1
×u
1
+m
2
×u
2
=m
1
×v
1
+m
2
×v
2
Block in the values,
\rightarrow→ \bold{\sf{m_1\:\times\:0\:+\:m_2\:\times\:0\:=\:3\:\times\:16\:+\:6\:\times\:v_2}}m
1
×0+m
2
×0=3×16+6×v
2
\rightarrow→ \bold{\sf{0=48+\:6\:\times\:v_2}}0=48+6×v
2
\rightarrow→ \bold{\sf{\:-\:48=\:6\:\times\:v_2}}−48=6×v
2
\rightarrow→ \bold{\sf{\dfrac{-48}{6}}\:=\:v_2}
6
−48
=v
2
\rightarrow→ \bold{\sf{-8\:=\:v_2}}−8=v
2
Now, we can calculate the kinetic energy of mass 6 kg using the formula.
FoRmUlA :
\bold{\huge{\boxed{\tt{\red{K.E\:=\:{\dfrac{1}{2}\:mv^2}}}}}}
K.E=
2
1
mv
2
Since we are calculating the kinetic energy of mass 6 kg i.e \bold{m_2}m
2
,we will consider the value of \bold{v_2}v
2
i.e - 8 instead of v.
Block in the values,
\rightarrow→ \bold{\sf{K.E\:=\:{\dfrac{1}{2}\:\times\:6\:\times\:8^2}}}K.E=
2
1
×6×8
2
\rightarrow→ \bold{\sf{K.E\:=\:{\dfrac{1}{2}\:\times\:6\:\times\:64}}}K.E=
2
1
×6×64
\rightarrow→ \bold{\sf{K.E\:=\:{\dfrac{384}{2}}}}K.E=
2
384
\rightarrow→ \bold{\sf{K.E\:=\:192}}K.E=192
•°• Kinetic Energy Of Mass 6 kg is 192 Joules.