Ball A of mass 1 ,kg moving along a straight | vedclass

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

Question

(A) For ball $A$: $m_A = 1 \,kg$, $u_A = 4 \,ms^{-1}$. For ball $B$: $m_B = 3 \,kg$, $u_B = 0 \,ms^{-1}$.Total initial momentum $p_i = m_A u_A + m_B u

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(A) For ball $A$: $m_A = 1 \,kg$, $u_A = 4 \,ms^{-1}$. For ball $B$: $m_B = 3 \,kg$, $u_B = 0 \,ms^{-1}$.Total initial momentum $p_i = m_A u_A + m_B u_B = (1 	imes 4) + (3 	imes 0) = 4 \,kg \cdot ms^{-1}$.After collision, both bodies stick together and move with a common velocity $v$.Total final momentum $p_f = (m_A + m_B)v = (1 + 3)v = 4v$.By the law of conservation of linear momentum, $p_i = p_f$, so $4 = 4v$, which gives $v = 1 \,ms^{-1}$.The force exerted on ball $B$ is the rate of change of momentum of ball $B$:$F_B = \frac{\Delta p_B}{\Delta t} = \frac{m_B(v - u_B)}{\Delta t} = \frac{3(1 - 0)}{0.1} = \frac{3}{0.1} = 30 \,N$.

Similar Questions

Discuss the completely inelastic collision in one dimension.

$A$ particle of mass $m_1$ undergoes a perfectly inelastic head-on collision with another particle of mass $m_2$ at rest $(m_2 > m_1)$. What fraction of the initial kinetic energy is converted into heat energy during this collision?

$A$ mass $m$ moving horizontally (along the $x-$axis) with velocity $v$ collides and sticks to a mass of $3m$ moving vertically upward (along the $y-$axis) with velocity $2v$. The final velocity of the combination is

$A$ completely inelastic collision is one in which the two colliding particles

Two particles $A$ and $B$ of equal mass $M$ are moving with the same speed $v$ as shown in the figure. They collide completely inelastically and move as a single particle $C$. The angle $\theta$ that the path of $C$ makes with the $X$-axis is given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module