Two bodies of masses 0.1 ,kg and 0.4 ,kg move | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) According to the law of conservation of linear momentum,the total momentum before collision equals the total momentum after collision.Let the mass

Vedclass · Accepted Answer

$1.2$

Vedclass · Answer

(D) According to the law of conservation of linear momentum,the total momentum before collision equals the total momentum after collision.Let the masses be $m_1 = 0.1 \,kg$ and $m_2 = 0.4 \,kg$,and their velocities be $v_1 = 1 \,m/s$ and $v_2 = -0.1 \,m/s$ (since they move towards each other).Let $V$ be the velocity of the combined mass after collision.$m_1 v_1 + m_2 v_2 = (m_1 + m_2) V$$(0.1)(1) + (0.4)(-0.1) = (0.1 + 0.4) V$$0.1 - 0.04 = 0.5 V$$0.06 = 0.5 V$$V = \frac{0.06}{0.5} = 0.12 \,m/s$The distance traveled by the combined mass in $t = 10 \,s$ is:$d = V 	imes t = 0.12 \,m/s 	imes 10 \,s = 1.2 \,m$.

Two bodies of masses $0.1 \,kg$ and $0.4 \,kg$ move towards each other with the velocities $1 \,m/s$ and $0.1 \,m/s$ respectively. After collision,they stick together. In $10 \,s$,the combined mass travels .......... $m$.

Similar Questions

Which of the following complexes is not a chelate?

The technique used to make numerous copies of a specific segment of $DNA$ quickly and accurately is

If $y=\sin (\log _e x)$,then $x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}$ is equal to

If $z=x+iy$ is a complex number satisfying $\left|z+\frac{i}{2}\right|^2=\left|z-\frac{i}{2}\right|^2$,then the locus of $z$ is

If $y=a \cos (\log x)+b \sin (\log x)$,where $a, b$ are parameters,then $x^2 y^{\prime \prime}+x y^{\prime}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module