Question

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$\bf\red{Two objects, each of mass 8.8 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 6m s-² before the collision during which they stick together. ...What will be the velocity of the combined object after collision????}$

$\large\bf{Note :-}$

$\bf\blue{Step by step Answer is needed...}$
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$\bf\blue{Question from NTSE ...}$
$\bf\blue{Asked by ltzBrainlyShivam....}$


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Answer 1

Given :-

Two objects, each of mass 8.8 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 6m s-²

To Find :-

What will be the velocity of the combined object after collision?

Solution :-

Here,

m₁ = m₂

And

u₁ = 6 m/s

u₂ = -6 m/s

Now

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(8.8)(6) + (8.8)(-6) = 8.8v₁ + 8.8v₂

52.8 + (-52.8) = 17.6v₁

52.8 - 52.8 = 17.6v₁

0 = 17.6v₁

Therefore

The velocity is 0

Answer 2

question

• Two objects, each of mass 8.8 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 6m s-² before the collision during which they stick together. ...What will be the velocity of the combined object after collision?

$\sf\huge\blue{\underline{ \underline{SOLUTION : } - - }}$

$\sf\huge\red{\underline{ \underline{given \: that : } - - }}$

$\sf \: ●Mass \: of \: one \: of \: the \: objects, \: m _ 1 = 8.8kg \\ \\ \sf●Mass \: of \: the \: other \: object, \: m _ 2 = 8.8 kg$

$\sf \: ●Velocity \: of \: m_1 \: before collision \: , \: u_1 = 6 m/s \\ \\ \sf \: ●Velocity \: of \: m_2, \: moving \: in opposite \\ \sf direction \: before collision, \: u_2 \:  = -6 m/s$

Note : - u2 is negative since it is in opposite directions of u1

$\sf\huge\blue{\underline{ \underline{Find \: out : } - - }}$

●The velocity of the combined object after collision

SOLUTION:

After collision, the objects stick together

$\qquad \red{ \underline{ \boxed{ \sf \: \therefore combined \: mass = m_1 + m_2}}}$

putting value in formula

$\sf \: \qquad \: combined \: mass =8.8 + 8.8 \\ \\ \\ \sf combined \: mass = 17.6kg$

finding momentum before after collision

Let v be the velocity of the combined object after the collision. By the law of conservation of momentum,

momentum before

☆ momentum = momentum of objects 1 + momentum of objects 2

$\longmapsto \sf \: Velocity = m_1u_1 + m_2u_2 \\ \\ \\ \longmapsto \sf \: Velocity = 8.8 \times 6 + 8.8 \times ( - 6) \\ \\ \\ \longmapsto \sf \: Velocity = 52.8 + ( - 52.8) \\ \\ \\ \longmapsto \sf \: Velocity = 52.8 - 52.8 \\ \\ \\ \longmapsto \sf \: \underline{\boxed{ \green{ \mathcal{Velocity = 0}}}}$

momentum after

$\longmapsto \sf \: momentum = total \: mass + velocity \\ \\ \\ \longmapsto \sf \: momentum =17.6 + v \\ \\ \\ \longmapsto \sf \: momentum =17.6v$

According to law of conservation of momentum,

Total momentum before collision = Total momentum after collision

$\longmapsto\qquad\qquad\sf0 = 17.6v \\ \\ \\ \longmapsto\qquad\qquad\sf \: v = \frac{0}{17.6} \\ \\ \\ \longmapsto\qquad\qquad\mathcal{v =0 m/s }$

Thus, the velocity of the combined object after the collision is 0 m/s