Question

A bullet with mass m = 5.21 g is moving horizontally with a speed v = 443 m/s when it strikes a block of hardened steel with mass M = 14.8 kg (initially at rest). The bullet bounces off the block in a perfectly elastic collision. (a) What is the speed (m/s) of the block immediately after the collision?

Answer 1

$V_{f2} =0.29 \dfrac{m}{s}$ :Speed of the block immediately after the collision.

What is the elastic collision?

From elastic collision theory.

Momentum of a body is given as

$P=m\times v$

where

p:Linear momentum

m: mass

v:velocity

There are 3 cases of collisions : elastic, inelastic and plastic.

For elastic collision the change in momentum remains constant

$P_{i} =P_{f}$

Nomenclature and data

m₁: ball mass= 5.21 g= 5.21*10⁻³kg

V₀₁: initial ball speed,  =412 m/s

$V_{f1}$: final ball speed

m₂: block mass =  14.8 kg  

V₀₂: initial block speed, = 0

$V_{f2}$: final block speed

We apply furmula

$P_{i} =P_{f}$

m₁$\times$V₀₁+m₂$\times$V₀₂=m₁$\times$Vf₁+m₂$\times$Vf₂

5.21$\times$10⁻³$\times$412+14.8$\times$0= 5.21$\times$10⁻³$\times$Vf₁+14.8$\times$Vf₂

2.15= 5.21*10⁻³$\times$Vf₁+14.8$\times$Vf₂ Equation (1)

For perfectly elastic collision the coefficient of elastic restitution (e) is equal to 1, and e is defined like this:

$e=\dfrac{v_{f2-V_{f1} } }{v_{01-V_{ 02} } }$

1*(V₀₁-V₀₂) =Vf₂-Vf₁  , V₀₂=0, V₀₁ =412 m/s

412=Vf₂-Vf₁

Vf₁=Vf₂-412 Equation (2)

We replace Equation (2) in Equation (1)

2.15= 5.21$\times$10⁻³(Vf₂-412)+14.8$\times$Vf₂

2.15= 5.21$\times$10⁻³*Vf₂-2.15+14.8*Vf₂

4.3=14.805Vf₂

$Vf_{2} =\dfrac{4.3}{14.805} =0.29 \dfrac{m}{s}$: (+) ,with equal direction  of the movement of the ball before the collision.

Hence    $V_{f2} =0.29 \dfrac{m}{s}$   Speed of the block immediately after the collision.

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