A bullet of mass 0.012 kg and horizontal speed 70 m s⁻¹ strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also estimate the amount of heat produced in the block.
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Hii dear,
◆ Answer-
h = 20.8 cm
H.E. = 28.54 J
◆ Explaination-
● Given-
m = 0.012 kg
M = 0.4 kg
u = 70 m/s
u' = 0
Final speed of system = ν
Applying the law of conservation of momentum:
mu + Mu' = (m+M) v
0.012×70 + 0.4×0 = (0.012+0.4)v
∴ v = 0.84/0.412
v = 2.04 m/s
Consider h be the height up to which the system rises.
Applying the law of conservation of energy to this system-
Potential energy (highest) = Kinetic energy (lowest)
(m+M)gh = 1/2 (m+M)v^2
∴ h = 1/2 v^2/g
h = 0.5×(2.04)^2/10
h = 0.208 m
The wooden block will rise to a height of 0.208 m.
Heat produced = Kinetic energy (bullet) – Kinetic energy (system)
H.E. = 1/2 mu^2 – 1/2 (m+M)v^2
H.E. = 0.5×0.012×(70)^2 – 0.5×0.412 × (2.04)^2
H.E. = 29.4–0.857
H.E. = 28.54 J
Hope this helps you...
◆ Answer-
h = 20.8 cm
H.E. = 28.54 J
◆ Explaination-
● Given-
m = 0.012 kg
M = 0.4 kg
u = 70 m/s
u' = 0
Final speed of system = ν
Applying the law of conservation of momentum:
mu + Mu' = (m+M) v
0.012×70 + 0.4×0 = (0.012+0.4)v
∴ v = 0.84/0.412
v = 2.04 m/s
Consider h be the height up to which the system rises.
Applying the law of conservation of energy to this system-
Potential energy (highest) = Kinetic energy (lowest)
(m+M)gh = 1/2 (m+M)v^2
∴ h = 1/2 v^2/g
h = 0.5×(2.04)^2/10
h = 0.208 m
The wooden block will rise to a height of 0.208 m.
Heat produced = Kinetic energy (bullet) – Kinetic energy (system)
H.E. = 1/2 mu^2 – 1/2 (m+M)v^2
H.E. = 0.5×0.012×(70)^2 – 0.5×0.412 × (2.04)^2
H.E. = 29.4–0.857
H.E. = 28.54 J
Hope this helps you...
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