
A 10g bullet moving at a speed of 200m/s stops after penetrating 5cm of wooden plank. The average speed force exerted on the bullet will be?
A) 2000N
B)-2000N
C) 5000N
D) -4000N
Answer
219k+ views
Hint: When the bullet starts to penetrate it loses its energy in doing so and hence its velocity will decrease and it will stop due to resistance provided by the wooden plank.
Complete step by step answer:
Given mass =10g
Converting gram into kilogram
m =10/1000 kg = 0.01kg Initial speed, u = 200m/s Final velocity, v = 0 m/s Distance covered (s) = 5cm
Converting centimeter to metre:
\[ \Rightarrow s = 0.05m\]
Here we can apply an equation of motion in order to calculate the deceleration (retarding acceleration) that acts on the bullet. The equation is:
${v^2} - {u^2} = 2 \times a \times s$
Let the deceleration (retarding acceleration) = - a
Minus sign cancels out on both sides
\[40000 = 2 \times a \times 5/100\]
Solving for (a) we get, \[40000 = a/10\] ∴$a = 400000m/{s^2}$ From Newton's Second law we know that force experienced by the bullet is equal to the product of its mass and acceleration. $F = m \times a$ = 0.01\[ \times \](-400000) N = - 4000N
∴\[F = - {\text{ }}4000{\text{N}}\]
Here the negative sign shows a force of 4000N acts in the opposite direction of the motion of the bullet.
$\therefore$ the correct option is D.
Note: For every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal and opposite force on object A.
This is Newton's Third Law of Motion. Also note that here we assume that the force exerted uniform(the average force) gives constant acceleration and hence we can use the kinematics equation for constant acceleration.
Complete step by step answer:
Given mass =10g
Converting gram into kilogram
m =10/1000 kg = 0.01kg Initial speed, u = 200m/s Final velocity, v = 0 m/s Distance covered (s) = 5cm
Converting centimeter to metre:
\[ \Rightarrow s = 0.05m\]
Here we can apply an equation of motion in order to calculate the deceleration (retarding acceleration) that acts on the bullet. The equation is:
${v^2} - {u^2} = 2 \times a \times s$
Let the deceleration (retarding acceleration) = - a
Minus sign cancels out on both sides
\[40000 = 2 \times a \times 5/100\]
Solving for (a) we get, \[40000 = a/10\] ∴$a = 400000m/{s^2}$ From Newton's Second law we know that force experienced by the bullet is equal to the product of its mass and acceleration. $F = m \times a$ = 0.01\[ \times \](-400000) N = - 4000N
∴\[F = - {\text{ }}4000{\text{N}}\]
Here the negative sign shows a force of 4000N acts in the opposite direction of the motion of the bullet.
$\therefore$ the correct option is D.
Note: For every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal and opposite force on object A.
This is Newton's Third Law of Motion. Also note that here we assume that the force exerted uniform(the average force) gives constant acceleration and hence we can use the kinematics equation for constant acceleration.
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