Question

An unloaded car moving with velocity u on a frictionless road can be stopped in a distance s. If the passengers add 40% to its weight and breaking force remains the same the stopping distance at velocity u is now

Answer 1

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➡ Given:

✏ An unloaded car moving with velocity u in a frictionless road can be stopped in a distance s.

➡ To Find:

✏ If the paasengers add 40% to its weight and bteaking force remains same then what will be stopping distance ?

➡ Formula derivation:

$\implies \rm \: {v}^{2} - {u}^{2} = 2as$

✏ v denotes final velocity

✏ u denotes initial velocity

✏ a denotes acceleration

✏ s denotes stopping distance

✏ in this case, v = 0

✏ a is negative.

$\implies \rm \: {0}^{2} - {u}^{2} = - 2as \\ \\ \implies \rm - {u}^{2} = - 2as \\ \\ \implies \rm \: \boxed{ \bf{\red{s = \frac{ {u}^{2} }{2a} }}}$

✏ As per ⬆⬆⬆ formula it is clear that...

✏ Stopping distance doesn't depend on weight of body.

✏ Therefore, stopping distance will remain same.

Answer 2

Complete question:

A man of mass m is standing on a stationary flat car of mass M. The car can move without friction along horizontal rails. The man starts walking with velocity v relative to the car. Work done by him

(a) is greater than 1/2 mv² if he walks along rails.

(b) is less than 1/2 mv² is he walks along rails.

(c) is equal to 1/2 mv² is he walks normal to rails.

(d) can never be less than 1/2 mv²

Answer:

The work done by him (b) is less than 1/2 mv² is he walks along rails and (c) is equal to 1/2 mv² is he walks normal to rails.

Explanation:

When the man starts walking along the rail, the man gains V velocity.

On applying conservation of momentum, we get,

MV = m(v - V)

∴ V = mv/(m + M)

The kinetic energy is provided to man and the car, thus, the kinetic energy is:

W = 1/2 m(v - V)² + 1/2mV²

W = 1/2 (mM/(m + M)) v²

The value of (mM/(m + M)) is less than m and M. Thus, work done is less than 1/2 mv². So, option (b) is correct.

When the man moves normal to the rail with velocity v, car will not move.

The kinetic energy of the man is equal to 1/2 mv². Thus, work done is is equal to 1/2 mv². So, option (c) is correct.