Question

small sphere of mass 2 kg moving with velocity u is equal to 4 ICAP - 7 J cap metre per second collides with smooth wall and returns with velocity V is equal to minus ICAP + 3 J cap. The magnitude of Impluse received by ball is​

Answer 1

Question:-

A small sphere of mass 2 kg moving with velocity $\sf{\overrightarrow{\sf{u}}=(4\ \hat i-7\ \hat j)\ m\ s^{-1}}$ collides with smooth wall and returns with velocity $\sf{\overrightarrow{\sf{v}}=(\hat i+3\ \hat j)\ m\ s^{-1}.}$ Find the magnitude of impulse received by ball.

Solution:-

• $\sf{m=2\ kg}$

• $\sf{\overrightarrow{\sf{u}}=(4\ \hat i-7\ \hat j)\ m\ s^{-1}}$

• $\sf{\overrightarrow{\sf{v}}=(\hat i+3\ \hat j)\ m\ s^{-1}.}$

Impulse received by the ball is obtained by its change in linear momentum.

$\longrightarrow\sf{\overrightarrow{\sf{I}}=\overrightarrow{\sf{\Delta p}}}$

$\longrightarrow\sf{\overrightarrow{\sf{I}}=m(\overrightarrow{\sf{v}}-\overrightarrow{\sf{u}})}$

$\longrightarrow\sf{\overrightarrow{\sf{I}}=2\left[(\hat i+3\ \hat j)-(4\ \hat i-7\ \hat j)\right]}$

$\longrightarrow\sf{\overrightarrow{\sf{I}}=2\left[\hat i+3\ \hat j-4\ \hat i+7\ \hat j\right]}$

$\longrightarrow\sf{\overrightarrow{\sf{I}}=2\left[-3\ \hat i+10\ \hat j\right]}$

$\longrightarrow\sf{\overrightarrow{\sf{I}}=(-6\ \hat i+20\ \hat j)\ kg\ m\ s^{-1}}$

Then the magnitude of impulse imparted is,

$\longrightarrow\sf{\left|\overrightarrow{\sf{I}}\right|=\sqrt{(-6)^2+20^2}}$

$\longrightarrow\sf{\left|\overrightarrow{\sf{I}}\right|=\sqrt{36+400}}$

$\longrightarrow\sf{\left|\overrightarrow{\sf{I}}\right|=\sqrt{436}}$

$\longrightarrow\sf{\underline{\underline{\left|\overrightarrow{\sf{I}}\right|=2\sqrt{109}}}}$

$\longrightarrow\sf{\underline{\underline{\left|\overrightarrow{\sf{I}}\right|=20.88}}}$

Answer 2

here, we have to use formula,

impulse = change in momentum

impulse = Pf - Pi

impulse = m(Vf - Vi)

here, m = 2 kg

Vf = (-i + 3j ) m/s

Vi = (4i - 7j) m/s

now,

impulse = 2 × ( -i + 3j - 4i + 7j)

= 2 × ( -5i + 10j)

=(-10i + 20j ) kg.m/s

hence, magnitude of impulse = √(10²+20²)

= 10√5 N.s