Question

A bullet of mass 20 g, travelling at a speed of 350 ms-1, strikes a steel plate at an angle of 30º with the plane of the plate. It ricochets off at the same angle, at a speed of 320 ms-1. What is the magnitude of the impulse that the steel plate gives to the projectile? If the collision with the plate takes place over a time ∆t = 10-3 s, what is the average force exerted by the plate on the bullet?

Answer 1

Answer on Question 65453, Physics, Mechanics, Relativity

Question:

A bullet of mass $20g$ , travelling at a speed of $350~ms^{-1}$ , strikes a steel plate at an angle of $30{}^{\circ}$ with the plane of the plate. It ricochets off at the same angle, at a speed of $320~ms^{-1}$ . What is the magnitude of the impulse that the steel plate gives to the projectile? If the collision with the plate takes place over a time $\Delta t = 10^{-3}s$ , what is the average force exerted by the plate on the bullet?

Solution:

a)

Let's first find $x$ - and $y$ -components of the impulse that the steel plate gives to the projectile (we take the positive $x$ - and $y$ -directions as in the picture above):

$$\begin{array}{l} J _ {x} = \Delta p _ {x} = m v _ {f x} - m v _ {i x} = m \left(- v _ {f x} \sin \alpha - v _ {i x} \sin \alpha\right) = \\ = 0. 0 2 k g \cdot \left(- 3 2 0 m s ^ {- 1} \cdot \sin 3 0 {}^ {\circ} - 3 5 0 m s ^ {- 1} \cdot \sin 3 0 {}^ {\circ}\right) = - 6. 7 N \cdot s, \\ \end{array}$$ $$\begin{array}{l} J _ {y} = \Delta p _ {y} = m v _ {f y} - m v _ {i y} = m \left(v _ {f y} \cos \alpha - v _ {i y} \cos \alpha\right) = \\ = 0. 0 2 k g \cdot \left(3 2 0 m s ^ {- 1} \cdot c o s 3 0 {}^ {\circ} - 3 5 0 m s ^ {- 1} \cdot c o s 3 0 {}^ {\circ}\right) = - 0. 5 2 N \cdot s. \\ \end{array}$$

We can find the magnitude of the impulse that the steel plate gives to the projectile from the Pythagorean theorem:

$$J = \sqrt {J _ {x} ^ {2} + J _ {y} ^ {2}} = \sqrt {(- 6 . 7 N \cdot s) ^ {2} + (- 0 . 5 2 N \cdot s) ^ {2}} = 6. 7 N \cdot s.$$

b) We can find the average force exerted by the plate on the bullet from the definition of the impulse:

$$J = \bar {F} \Delta t,$$ $$\bar {F} = \frac {J}{\Delta t} = \frac {6.7 \, N \cdot s}{10^{-3} \, s} = 6700 \, N.$$

Answer:

a) $J = 6.7 \, N \cdot s$.

b) $\bar{F} = 6700 \, N$.

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