Question

A boy of 50 kg mass is running with a velocity of 2 m/s. He jumps over a stationary cart of 2 kg while running. Find the velocity of cart after the boy jumps onto it.

Answer 1

Answer:

here

Explanation:

$m1v1 = m2v2$

law of conservation of momentum

50×2=52×v2

100/52=v2

v2=1.92m/s

Answer 2

$\bf{\underline{\underline{Answer:}}}$

$\bf{\therefore 1.92\:m\:s^{-1} }$

$\bf{\underline {Given:}}$

$⟹Initial \:velocity \:of \:boy, m_1=50\:kg$

$⟹Mass \:of\: cart, \:m_2=2\:kg$

$⟹Initial\: velocity\: of\: cart,\: u_2=0$

$\bf{\underline {To\:Find:}}$

$⟹Final\: velocity \:of \:cart, \:v_2=\:?$

Since, the boy jumps onto the cart, the final velocity, $v_1$ of the boy will be equal to that of the cart.

Therefore$, v_1=v_2$

$\bf{\underline{\underline{Step\: by\: step \:explanation:}}}$

We know that,

$m_1u_1+m_2m_2=m_1v_1+m_2v_2$

Substituting the value, we get

$50×2+2×0=50 ×v_1+2×v_2$

$100=50×v_2+2×v_2$ $[v_1=v_2]$

$100=v_2×52$

$v_2=\dfrac{100}{52}=\bf 1.92\:m\:s^{-1}$

Therefore, the velocity of the cart after the boy jumps onto it is equal to $\bf{1.92\:m\:s^{-1}}$ .Since velocity has positive sign, the cart will go in the same direction in which the boy was running.