The weight of the buggy was 40n on mars. when the buggy landed on mars it rested on an area of 0.025m squared. calculate the pressure exerted by the buggy on the surface of mars. give the unit

The speed of m2 is 0.6 m/s and its direction is to the right.

Explanation:

This numerical can be solved easily by applying law of conservation of momentum to it. According to law of conservation of momentum:

Total Momentum Before Collision = Total Momentum After Collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where,

m₁ = Mass of 1st air glider = 0.25 kg

m₂ = Mass of 2nd air glider = 0.5 kg

u₁ =  Speed of 1st air glider before collision = 0.9 m/s

u₂ = Speed of 2nd air glider before collision = 0 m/s (at rest)

v₁ =  Speed of 1st air glider after collision = - 0.3 m/s (negative sign due to change in direction of velocity)

v₂ = Speed of 2nd air glider after collision = ?

Therefore,

(0.25 kg)(0.9 m/s) + (0.5 kg)(0 m/s) = (0.25 kg)(-0.3 m/s) + (0.5 kg)v₂

0.225 kg.m/s + 0.075 kg.m/s = (0.5 kg)v₂

v₂ = (0.3 kg.m/s)/(0.5 kg)

v₂ = 0.6 m/s

Positive sign indicates that v₂ is directed towards right