1. A force of 600 dynes acts on a glass ball of mass 200 g for 12 s. If initially the ball is at rest,
find (1) Final velocity (11) Distance covered.
2. A bullet of mass 30 g, and moving with a velocity * hits a wooden target with a force of
187.5 N. If the bullet penetrates 80 cm, find the value of x.
3. A car of mass 1000 kg develops a force of 500 N over a distance of 49 m. If initially the car is
at rest find (i) Final velocity (u) Time for which it accelerates.
PLEASE HELP ME...I NEED IT URGENTLY....THANK YOU
Answers
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★ Sᴏʟᴜᴛɪᴏɴ 1 :-
Given ,
- Force = 600 dynes
- Initially at rest
- Mass = 200 g
- Time = 12s
We need to find ,
- i) Final velocity
- ii) 2nd velocity
i) sol :-
As we know that
F = m (v - u/t)
→ 600 = 200 (v - 0/12)
→ 600 - 200 = v/12
→ 400 × 12 = v
→ v = 4800 cm/s
ii) sol:-
To find distance , firstly we need to find acceleration
a = v - u/t
• a = 4800 - 0/t
• a = 4800/12
• a = 400 cm/s²
Now , finding distance using 3rd equation of motion
S = ut + 1/2 at²
→ S = 0(12) + 1/2 (400)(12)²
→ S = 200 × 144
→ S = 28,800 cm
∴ v = 4800 cm/s & s = 28800 cm
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★ Sᴏʟᴜᴛɪᴏɴ 2 :-
Given ,
- Mass of bullet 30g = 0.03 kg
- Initial Velocity = x
- Force = 187.5 N
- Penetrates = 80cm = 0.8 m
We need to find ,
- x = ?
Using formula
F = m × ( -a)
Note :- Here , the acceleration will be negative because the bullet penetrates , that means go back 80 cm , that means it's is retardation.
• - a = 187.5 ÷ 0.03
• - a = 6250
• a = - 6250 m/s²
Here , the final Velocity will be zero because the bullet hits the wooden target .
Now, finding x , using second equation of motion.
v² - u² = 2aS
→ 0² - (x)² = 2(-6250)(0.8)
→ - x² = 8/5 × (-6250)
→ - x² = - 10000
→ x² = 10000
→ x = √10000
→ x = 100 m/s
∴ x = 100 m/s
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★ Sᴏʟᴜᴛɪᴏɴ 3 :-
Given,
- Mass of car = 1000 kg
- Force = 500 N
- Distance = 49m
- Initially at rest
We need to find ,
- Final velocity
- Time for which it accelerates
i) sol :-
As we know that
F = ma
→ 500 = 1000 × a
→ a = 500/1000
→ a = 0.5 m/s²
Now, finding final velocity using 2nd equation of motion.
v² - u² = 2as
→ v² - 0² = 2(1/2)(49)
→ v² = 49
→ v = √49
→ v = 7 m/s
ii) sol :-
Now , finding time using first equation of motion
v = u + at
→ 7 = 0 + (1/2)t
→ 7 = t/2
→ 7 × 2 = t
→ t = 14s
∴ v = 7 m/s & t = 14s