The change of momentum in each ball of mass 60 gm, moving in opposite ...
Before collision :---> both ball comes closer with velocity of u1 vector in i cap and u2 vector in -i cap respectively
After collision :--> both balls go far away with velocity of v1 vector in -i cap and v2 vector in i cap respectively
And both balls have same masses = 0.06 kg
Now, u1=4 i cap , u2= -4 i cap , v1= -4 i cap , v2=4 i cap
So,
Impulse imparted on each ball = change in momentum of each ball
Therefore, delta P = m (v1 vector - u1 vector)
delta P = 0.06 (-4 i cap - 4 i cap)
delta P = 0.48 kg m/s ...ans...
$$Hope it's help... $$
The change of momentum in each ball of mass 60 gm, moving in opposite ...
Given:
Mass of each ball, m = 60 g = 0.06 kg
Initial velocity of the first ball, u1 = 4 m/s
Initial velocity of the second ball, u2 = -4 m/s (opposite direction)
Final velocity of both the balls after collision, v1 = v2 = 4 m/s (rebound with same speed)
To find: Change in momentum of each ball
Solution:
1. Momentum before collision:
Momentum of the first ball, p1 = m × u1 = 0.06 × 4 = 0.24 kg m/s (to the right)
Momentum of the second ball, p2 = m × u2 = 0.06 × (-4) = -0.24 kg m/s (to the left)
2. Momentum after collision:
Since the balls collide and rebound with the same speed, the direction of momentum for each ball gets reversed.
Momentum of the first ball, p1' = m × v1 = 0.06 × 4 = 0.24 kg m/s (to the left)
Momentum of the second ball, p2' = m × v2 = 0.06 × (-4) = -0.24 kg m/s (to the right)
3. Change in momentum:
The change in momentum is given by the difference between momentum before and after collision.
Change in momentum of the first ball, Δp1 = p1' - p1 = -2 × 0.24 = -0.48 kg m/s (to the left)
Change in momentum of the second ball, Δp2 = p2' - p2 = 2 × 0.24 = 0.48 kg m/s (to the right)
4. Total change in momentum:
The total change in momentum for the system of both balls is the sum of individual changes in momentum.
Δp = Δp1 + Δp2 = -0.48 + 0.48 = 0 kg m/s
Therefore, the correct option is (c) 0.48 kg-m/s.
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