Question

an object of mass 1 kg travelling in a straight line with the velocity of 10m/s collides with an sticks to a stationary wooden vlog of mass 5kg then they both move together in a same straight line. Calculate the total momentum & combine velocity???



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Answer 1

Answer:

Mass of the object, m1 = 1 kg

Velocity of the object before collision, v1 = 10 m/s

Mass of the stationary wooden block, m2 = 5 kg

Velocity of the wooden block before collision, v2 = 0 m/s

Total momentum before collision = m1 v1 + m2 v2

= 1 (10) + 5 (0) = 10 kg m sˆ’1

It is given that after collision, the object and the wooden block stick together.

Total mass of the combined system = m1 + m2

Velocity of the combined object = v

According to the law of conservation of momentum:

Total momentum before collision = Total momentum after collision

m1 v1 + m2 v2 = (m1 + m2) v

1 (10) + 5 (0) = (1 + 5) v

v = 10 / 6

= 5 / 3

The total momentum after collision is also 10 kg m/s.

Total momentum just before the impact = 10 kg m sˆ’1

Total momentum just after the impact = (m1 + m2) v = 6 — 5 / 3 = 10 kg ms-1

Hence, velocity of the combined object after collision = 5 / 3 ms-1

Explanation:

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Answer 2

Solution :-

• m = 10g = $\dfrac{10}{1000}$ = 0.01kg

• u = 150$\dfrac{m}{s}$

• v = 0$\dfrac{m}{s}$

• t = 0.03 seconds

We know :-

⠀⠀⠀⠀⟹${\boxed{\large{\bold{v\:=\:u\:+at}}}}$

⠀⠀⠀⠀⟹ 0 = 150 + a(0.03)

⠀⠀⠀⠀⟹ a = $\dfrac{-150}{0.03}$

⠀⠀⠀⠀⟹ - 5000 $\dfrac{m}{s}^{2}$

• S = ?

• F = ?

⠀⠀⠀⠀⟹${\boxed{\large{\bold{{v}^{2}\:-\:{u}^{2}\:=\:2as}}}}$

⠀⠀⠀⠀⟹${{(0)}^{2}\:-\:{(150)}^{2}}$

⠀⠀⠀⠀⟹ 2 × 5000 × s

⠀⠀⠀⠀⟹ s = $\dfrac{150\:×\:150}{2\:×\:5000}$

⠀⠀⠀⠀⠀⠀⠀⟹ s = $\dfrac{22500}{10000}$

⟹ $\boxed {s\: =\:2.25m}$

⠀⠀∴ The penetration distance of the bullet in the wooden block

⠀⠀⠀⠀⠀⠀⠀= 2.25 m

Magnitude of force :-

⠀⠀⠀⠀⠀⠀⠀⠀⠀${\boxed{\large{\bold{f\:=\:ma}}}}$

⠀⠀⠀⠀⠀⠀⠀⠀⟹ $\dfrac{10}{1000}$ × $5000$

⠀⠀⠀⠀⠀⠀⠀⟹ Force = 50 N