Question

plz do it fast with method​

Answer 1

$\huge\underline\blue{\sf Answer:}$

$\red{\boxed{\sf Maximum\: Height\:(H_{max})=1.25\:cm }}$

$\huge\underline\blue{\sf Solution:}$

$\large\underline\pink{\sf Given: }$

• Initial mass (m) = 10g

• Initial Velocity (u) = 100$\sf{cm^{-1}}$

After Collision

• Mass (M) = 10 + 10 = 20g

• Let Velocity will be = v

$\large\underline\pink{\sf To\:Find: }$

• Maximum Height reached by the system ($\sf{H_{max}}$)=?

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By law of Conservation of Momentum :-

$\large{\boxed{\sf mu=Mv}}$

$\large\implies{\sf 10×100=20×v}$

$\large\implies{\sf v=\frac{10×100}{20}}$

$\large\implies{\sf v=50cms^{-1}}$

Now for Maximum Height we know :-

$\large{\boxed{\sf H_{max}=\frac{v^2}{2g}}}$

Here ,

g = 1000$\sf{cms^{-2}}$

$\large\implies{\sf H_{max}=\frac{(50)^2}{2×1000}}$

$\large\implies{\sf H_{max}=\frac{2.5}{2}}$

$\large\implies{\sf H_{max}=1.25\:cm}$

$\red{\boxed{\sf Maximum\: Height\:(H_{max})=1.25\:cm }}$

Hence,

Maximum Height reached by the system after Collision is 1.25cm

Answer 2

$\huge\mathfrak\red{Answer}$

FIRSTLY CONVERT ALL

THE UNITS AS FOLLOWS...

$mass \: of \: moving \: object = 10g \: \\ in \: kgs = 10 \times {10}^{ - 3} \\ mass \: of \: moving \: object = {10}^{ - 2} kg$

Mass of object(M) =0.01kg

$velocity \: of \: moving \: object = 100cm {sec}^{ - 1} \\ in \: m {sec}^{ - 1} = 100 \times {10}^{ - 2} m {sec}^{ - 1} \\ \\ velocity \: of \: moving \: object = 1m {sec}^{ - 1}$

Velocity of moving object(v) =1msec^-1

$mass \: of \: bob = 10g \\ in \: kgs = 10 \times {10}^{ - 3} kg \\ \\ mass \: of \: bob \: in \: kgs = {10}^{ - 2} kg$

Mass of bob(m) =0.01kg

VELOCITY OF BOB(u')

Mv + mu = Mu' + mu'...... [1]

Law of conservation of linear momentum

FINAL VELOCITY OF BOTH BOB AND MOVING OBJECT IS SAME BECAUSE THEY STICK WITH EACH OTHER.

PUT VALUES IN [1]

0.01(1) + 0.01(0) = 0.01(u') + 0.01(u')

(Initial velocity of the bob is 0)

0.01=0.01(2u')

0.01=0.02u'

0.5=u'

$now \: the \: final \: velocity \: of \: bob \: is \: 0.5m {sec}^{ - 1} \\ final \: mass \: of \: bob \: = 0.02 + 0.02 = 0.04kg \\ \\ apply \: law \: of \: conservation \: of \: energy \\ initial \: energy = \frac{1}{2} m {u}^{2} \\ \\let \: the \: height \: gain \: by \: bob \: is \: h \: than \\ final \: energy \: = mgh \\ \\ \frac{1}{2} m {u}^{2} = mgh \\ \frac{ {u}^{2} }{2} = gh \\ \\ (u = 0.5m {sec}^{ - 1} ) \\ \\ \frac{ {(0.5)}^{2} }{2} = 10(h) \\ \\ \frac{0.25}{20} = h \\ \\ 0.0125m = height \\ \\ in \: centimeters = 0.0125 \times {10}^{2} cm \\ \\ height = 1.25cm$