Two solid rubber balls A and B having masses 200 g and 400 g respectively are moving in opposite directions with velocity of A equal to 0.3 m/s. After collision the two balls come to rest when the velocity of B is

Question

Two solid rubber balls A and B having masses 200 g and 400 g respectively are moving in opposite directions with velocity of A equal to 0.3 m/s. After collision the two balls come to rest when the velocity of B is (A) 0.15 m/s (B) 1.5 m/s (C) -0.15 m/s (D) noneof the these

Tardigrade Admin · Accepted Answer

According to the law of conservation of linear momentum mA uA+mBuB=mAVA+mBVB ∴ (200 g)(0.3 m/s)+(400 g)(ub) =(200 g)(0)+(400 g)(0) ∴ uB=-((200 g)(0.3 m/ s)/(400 g )) =-0.15 m/s

Q. Two solid rubber balls $A$ and $B$ having masses $200 g$ and $400 g$ respectively are moving in opposite directions with velocity of A equal to $0.3 m / s$ . After collision the two balls come to rest when the velocity of $B$ is

$0.15 m / s$

$1.5 m / s$

$- 0.15 m / s$

noneof the these

According to the law of conservation of linear momentum
$m_{A} u_{A} + m_{B} u_{B} = m_{A} V_{A} + m_{B} V_{B}$
$∴ (200 g) (0.3 m / s) + (400 g) (u_{b})$
$= (200 g) (0) + (400 g) (0)$
$∴ u_{B} = - \frac{( 200 g ) ( 0.3 m / s )}{( 400 g )}$
$= - 0.15 m / s$

Q. Two solid rubber balls A and B having masses 200g and 400g respectively are moving in opposite directions with velocity of A equal to 0.3m/s. After collision the two balls come to rest when the velocity of B is

0.15m/s

1.5m/s

−0.15m/s