Answer:
As collision is elastic,thus we can use conservation of momentum equation
mA=0.2 kg
(vB)1=0 m/s.......................as it is on rest before collision
(vA)1=4 m/s
(vA)2=-1 m/s
(vB)2=2 m/s
using equation
(mA*vA+mB*vB)1= (mA*vA+mB*vB)2
Where 1 and 2 represents before and after collision
(0.2*4)+(mB*0)=(0.2*-1)+(mB*2)
0.8=-0.2+(2mB)
mass of object B=mB=0.3 Kg
Answer:

Explanation:
By Snell's law we know at the left surface




now we have


now on the other surface we know that
angle of incidence = 

so again we have

so we have


also we know that


By solving above equation we have

Answer:
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Explanation:
Given that,
Initial speed of the car, u = 88 km/h = 24.44 m/s
Reaction time, t = 2 s
Distance covered during this time, 
(a) Acceleration, 
We need to find the stopping distance, v = 0. It can be calculated using the third equation of motion as :


s = 74.66 meters
s = 74.66 + 48.88 = 123.54 meters
(b) Acceleration, 


s = 37.33 meters
s = 37.33 + 48.88 = 86.21 meters
Hence, this is the required solution.