A bullet having a mass of 10 g and moving with a speed of 1.5 m/s, penetrates a thick wooden plank of mass 900 g. The plank was initially at rest. The bullet gets embedded in
the plank and both move together.
Determine their velocity. Please explain also.
Correct answer will be marked brainliest!!!
Answers
Given :
▪ Mass of bullet = 10g
▪ Initial velocity = 1.5m/s
▪ Mass of wooden plank = 900g
▪ Bullet collides with wooden plank and gets embedded in the plank.
To Find :
▪ Final velocity of the combination.
(bullet + wooden plank)
Concept :
➳ Since, No net force acts on the whole system, we can easily apply concept of linear momentum conservation.
➳ Initial linear momentum of system = Final linear momentum of system
➳ Initial momentum of wooden plank = zero (i.e., at rest initially)
➳ Momentum is defined as the product of mass and velocity.
Calculation :
➠ (P)initial = (P)final
➠ mv + MV = (m + M)v'
where,
◕ m denotes mass of bullet
◕ v denotes velocity of bullet
◕ M denotes mass of wooden plank
◕ V denotes velocity of wooden plank
◕ v' denotes velocity of the combination
➠ (0.01×1.5) + (0.9×0) = (0.9+0.01)v'
➠ 0.015 + 0 = 0.91v'
➠ v' = 0.015/0.91
➠ v' = 0.016m/s
Given :
Mass of bullet (m1) = 10g = 10 = 1 = 0.01 kg
1000 100
Mass of wooden plank(m2) = 900 = 9 = 0.9 kg
1000 10
Initial speed of bullet ( velocity ) ( u1) = 1.5 m/s
Initial velocity of wooden plank ( u2) = 0 m/s
To find :
final velocity of bullet ( v1) =?
final velocity of wooden plank ( v2) = ?
Formula :
(m1×u1) + (m2×u2) = ( m1×v1 ) + (m1×v2)
Hint :
v1 = v2 = v
Solution :
(m1×u1) + ( m2×u2) = (m1×v1) + ( m2×v2)
( 0.01 × 1.5) + ( 0.9×0) = ( 0.01×v1) + ( 0.9×v2)
0.015 + 0 = 0.01v1 + 0.9v2
0.015 = v ( 0.01 + 0.9)
0.015 = v( 0.91)
v= 0.015
0.91
v = 0.02 m/s
But
if you give ,
m2 = 90 g = 0.09 kg
then,
(0.01×0.15) + (0.09×0) = ( 0.01×v1)+(0.09×v2)
0.015 +0 = v( 0.01+0.09)
0.015 = v( 0.10)
v= 0.015
0.10
v= 0.15 m/s