A bullet of mass 0.02 kg moving with a speed of 200 m/s strikes a 2 kg wooden block suspended by a 1m long thread and is embedded in the block. What is the maximum inclination of the thread with the vertical ?
answer=37 degree
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Initial conditions:
mass of bullet = 0.02kg
velocity of bullet = 200 m/s
mass of block = 2kg
velocity of block = 0
Just after strike:
mass of system = 2+0.02 = 2.02 kg
velocity = v
From momentum conservation equation
0.02×200 + 2×0 = 2.02×v
⇒ 2.02v = 4
⇒ v = 4/2.02
⇒ v = 1.98 m/s
After the strike, let the block with the bullet rises to a height H.
After strike,
KE of system = 0.5×2.02×1.98² = 3.96 J
PE of system = mg×0 = 0
Total Energy = 3.96+0 = 3.96J
At height H,
velocity = 0
KE = 0
PE = mgH = 2.02×9.8×H = 19.8H J
Total energy = 19.81H
Since energy is conserved,
19.81H = 3.96
⇒ H = 3.96/19.8
⇒ H = 0.2 m
If the inclination of the thread is Ф, (see the attachment)
cos Ф = 0.8/1
⇒ cos Ф = 0.8
⇒ Ф= cos⁻¹ 0.8
⇒ Ф = 37°
mass of bullet = 0.02kg
velocity of bullet = 200 m/s
mass of block = 2kg
velocity of block = 0
Just after strike:
mass of system = 2+0.02 = 2.02 kg
velocity = v
From momentum conservation equation
0.02×200 + 2×0 = 2.02×v
⇒ 2.02v = 4
⇒ v = 4/2.02
⇒ v = 1.98 m/s
After the strike, let the block with the bullet rises to a height H.
After strike,
KE of system = 0.5×2.02×1.98² = 3.96 J
PE of system = mg×0 = 0
Total Energy = 3.96+0 = 3.96J
At height H,
velocity = 0
KE = 0
PE = mgH = 2.02×9.8×H = 19.8H J
Total energy = 19.81H
Since energy is conserved,
19.81H = 3.96
⇒ H = 3.96/19.8
⇒ H = 0.2 m
If the inclination of the thread is Ф, (see the attachment)
cos Ф = 0.8/1
⇒ cos Ф = 0.8
⇒ Ф= cos⁻¹ 0.8
⇒ Ф = 37°
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