A bullet of mass 10 g moving horizontally at a speed of 50√7 m/s strikes a block of mass 490 g kept on a frictionless track as shown in figure (9-E17). The bullet remains inside the block and system proceeds towards the semicircular track of radius 0.2 m. Where will the block strike the horizontal part after leaving the semicircular track?
Answers
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ANSWER::
Mass of block = 490 gm
Mass of bullet = 10 gm
It is an elastic collision since, the bullet is embedded inside the block .
Initial velocity of block , v₁ = 50√7 m/s
Velocity of block , v₂ = 0
Let final velocity of both be v.
V₁ = Velocity of A block
V₂ = Velocity of B block
10 x 10⁻³ x 50 x √7 + 10⁻³ x 190 x 0 = (490 + 10) x 10⁻³ x V₁
V₁ = √7 m/s
When block losses contact at D the component mg will act on it.
m(V₂)² / r = mg sinФ
(V₂)² = rg sinФ [Equation 1]
Now applying work energy principle,
(1/2) m x (V₂)² - (1/2) x m x (V₁)² = - mg ( 0.2 + 0.2 sinФ)
(1/2) x gr sinФ - (1/2) x (√7)² = -mg (0.2 + 0.2 sinФ)
3.5 - (1/2) x 9.8 x 0.2 x sinФ = 9.8 x 0.2 ( 1 + sinФ)
3.5 - 0.98 sinФ = 1.96 + 1.96 sinФ
sin Ф = 1/2
Ф = 30°
Therefore , Angle of Projection = 90° - 30° = 60°
And , Time of reaching ground = √(2h/g) = √[2 x (0.2 + 0.2 x sin 30°) / 9.8]
= 0.247 seconds
Now , distance travelled in horizontal direction ,
S = V cos Ф x t
S = √(gr sinФ) x t
S = √(9.8 x 2 x 1/2) x 0.247 = 0.196 m
Therefore , total distance = (0.2 - 0.2 cos30°) + 0.196 = 0.22 m
Hope it helps!
