Perpendicular acceleration:
F = ma
a = 4 / 2 = 2 m/s²
Perpendicular distance:
s = ut + 1/2 at²
s = 0 x 4 + 1/2 x 2 x 4²
s = 16 m
Horizontal distance:
s = ut
= 3 x 4
= 12 m
Total distance = √(12² + 16²)
= 20 m.
Answer:
Speed of both blocks after collision is 2 m/s
Explanation:
It is given that,
Mass of both blocks, m₁ = m₂ = 1 kg
Velocity of first block, u₁ = 3 m/s
Velocity of other block, u₂ = 1 m/s
Since, both blocks stick after collision. So, it is a case of inelastic collision. The momentum remains conserved while the kinetic energy energy gets reduced after the collision. Let v is the common velocity of both blocks. Using the conservation of momentum as :



v = 2 m/s
Hence, their speed after collision is 2 m/s.
Well first graph represents rectangular hyperbola
vu = c^2 ( c is constant)
AS 1/v + 1/u = 1/f
Take1/ f to be constant c
1/v = c - 1/u
it is of the form y = - x + k
Slope = -1 having intercept k as shown in fig 2
Answer:
Work done gets doubled.
Explanation:
The work done by a force is given by :
W = Fd
Where
F is force and d is distance move
If the force is doubled and the distance moved remain the same, it would mean that the work done becomes double of the initial work done.