A small ball of mass 1 Kg moving with a velocity of 12 m/s undergoes a direct central impact with a stationary ball of mass 2 kg. The impact is perfectly elastic. The speed (in m/s) of 2 kg mass ball after the impact will be

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Solution

1. Conserving linear momentum
$1 \times 12 = 1 \times V_{1} + 2 \times V_{2}$
$12 = V_{1} + 2 V_{2} . . . (i)$
2. Velocity of approach = Velocity of separation
$12 - 0 = V_{2} - V_{1}$
$V_{2} - V_{1} = 12... (i i)$
From equalions (i) and (ii),we get
$V_{2} = 8 \frac{m}{s}$

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A small ball of mass 1 Kg moving with a velocity of 12 m/s undergoes a direct central impact with a stationary ball of mass 2 kg. The impact is perfectly elastic. The speed (in m/s) of 2 kg mass ball after the impact will be

1. Conserving linear momentum 1×12=1×V1+2×V2 12=V1+2V2...(i) 2. Velocity of approach = Velocity of separation 12−0=V2−V1 V2−V1=12...(ii) From equalions (i) and (ii),we get V2=8ms

1. Conserving linear momentum
$1 \times 12 = 1 \times V_{1} + 2 \times V_{2}$
$12 = V_{1} + 2 V_{2} . . . (i)$
2. Velocity of approach = Velocity of separation
$12 - 0 = V_{2} - V_{1}$
$V_{2} - V_{1} = 12... (i i)$
From equalions (i) and (ii),we get
$V_{2} = 8 \frac{m}{s}$