Question

A 100kg golf ball is moving to the right which a velocity of 20 meter per second .it makes a head on collision with 8 kg steel ball initially at rest .compute velocities of the ball after collision?

Answer 1

Answer:

AFTER COLLISION THE VELOCITIES OF THE BALL WILL BE AS THE FOLLOWING:-

GOLF BALL WILL COME AT 20 TO 0

STEEL BALL WILL COME AT 0 TO 50√2

Explanation:

m= 100 kg

v = 20m/s

K.E. = 1/2m*v*v = 0.5*100*20*20 = 20,000 N

SO THIS ENERGY WILL BE PASSED ONTO THE STEEL BALL AS PER THE LAW OF CONSERVATION OF ENERGY.

M = 8 KG

K1 = K2

20000 = 0.5*8*V*V

20000/4 = V*V

5000 = V²

V = √5000 = 50√2 M/s

DO MARK IT AS BRAINLIEST.

Answer 2

Answer:

• The Final velocity (v₁) of golf ball is 17 m/s.
• The Final velocity (v₂) of steel ball is 37 m/s.

Given:

$\boxed{\begin{minipage}{10 em} \textbf{\underline{Golf Ball}}\colon \\\\ \bullet \rm u_1 = 20 \;m/s \\\bullet \rm v_1 = ? \\\bullet \rm m_1 = 100\;Kg\end{minipage}}\quad\boxed{\begin{minipage}{10 em}\textbf{\underline{Steel Ball}}\colon \\\\ \bullet \rm u_2 = 0 \; m/s \\\bullet \rm v_2 = \; ? \\\bullet \rm m_2 = 8 \;Kg \end{minipage}}$

Explanation:

$\rule{300}{1.5}$

From the final velocity of one - dimensional collision of first body (m₁ = 100 kg),

$\large \bigstar\; \boxed{\tt v_1 = \Bigg(\dfrac{m_1 - m_2}{m_1 + m_2}\Bigg)u_1+\Bigg(\dfrac{2\;m_2}{m_1+m_2}\Bigg)u_2}$

$\mathfrak{Here}\begin{cases}\sf{m_1}\text{ Denotes Mass of 1st Body}\\\sf{m_2}\text{ Denotes Mass of 2nd Body}\\\sf{u_1}\text{ Denotes Initial velocity of 1st body}\\\sf{u_2}\text{ Denotes initial velocity of 2nd Body}\end{cases}$

Now,

$\large \boxed{\tt v_1 = \Bigg(\dfrac{m_1 - m_2}{m_1 + m_2}\Bigg)u_1+\Bigg(\dfrac{2\;m_2}{m_1+m_2}\Bigg)u_2}$

Substituting the values,

$\displaystyle\dashrightarrow \tt v_1=\Bigg(\dfrac{100 - 8}{100 + 8}\Bigg)\times20+\Bigg(\dfrac{2\times8}{100+8}\Bigg)\times0\\\\\\\dashrightarrow \tt v_1=\Bigg(\dfrac{100 - 8}{100 + 8}\Bigg)\times20+0\\\\\\\dashrightarrow\tt v_1=\Bigg(\dfrac{100 - 8}{100 + 8}\Bigg)\times20\\\\\\\dashrightarrow\tt v_1=\dfrac{92}{108}\times20\\\\\\\dashrightarrow\tt v_1 = \dfrac{1840}{108}\\\\\\\dashrightarrow\tt v_1 = \cancel{\dfrac{1840}{108}}\\\\\\\dashrightarrow\tt \large\underline{\boxed{\red{\tt v_1 \approx 17\;m/s}}}$

∴ The Final velocity of Golf ball is 17 m/s.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From the final velocity of one - dimensional collision of second body (m₂ = 8 kg),

$\large \bigstar\; \boxed{\tt v_2 = \Bigg(\dfrac{m_2 - m_1}{m_1 + m_2}\Bigg)u_2+\Bigg(\dfrac{2\;m_1}{m_1+m_2}\Bigg)u_1}$

$\mathfrak{Here}\begin{cases}\sf{m_1}\text{ Denotes Mass of 1st Body}\\\sf{m_2}\text{ Denotes Mass of 2nd Body}\\\sf{u_1}\text{ Denotes Initial velocity of 1st body}\\\sf{u_2}\text{ Denotes initial velocity of 2nd Body}\end{cases}$

Now,

$\large \boxed{\tt v_2 = \Bigg(\dfrac{m_2 - m_1}{m_1 + m_2}\Bigg)u_2+\Bigg(\dfrac{2\;m_1}{m_1+m_2}\Bigg)u_1}$

Substituting the values,

$\displaystyle\dashrightarrow \tt v_2=\Bigg(\dfrac{8 - 100}{100 + 8}\Bigg)\times0+\Bigg(\dfrac{2\times100}{100+8}\Bigg)\times20\\\\\\\dashrightarrow\tt v_2=\Bigg(\dfrac{2\times100}{100 + 8}\Bigg)\times20+0\\\\\\\dashrightarrow\tt v_2=\Bigg(\dfrac{2\times100}{100 + 8}\Bigg)\times20\\\\\\\dashrightarrow\tt v_2=\dfrac{200}{108}\times20\\\\\\\dashrightarrow\tt v_2=\dfrac{4000}{108}\\\\\\\dashrightarrow\tt v_2=\cancel{\dfrac{4000}{108}}$

$\dashrightarrow\tt \large\underline{\boxed{\red{\tt v_2\approx37\;m/s}}}$

∴ The Final velocity of Steel ball is 37 m/s.

$\rule{300}{1.5}$

$\rule{300}{1.5}$