Sa steel ball with mass m=5.21 g is moving horizontally with speed =412 m/s when it strikes a block of hardened steel with mass =14.8 kg (initially at rest). the ball bounces off the block in a perfectly elastic collision. (a) what is the speed of the block immediately after the collision?

Vf₂ = 0.29 m/s :Speed of the block immediately after the collision.

Explanation:

Theory of collisions

Linear momentum is a vector magnitude (same direction and direction as velocity) and its magnitude is calculated like this:

P=m*v

where

p:Linear momentum

m: mass

v:velocity

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

For the three cases the total linear momentum quantity is conserved:

P₀=Pf  Formula (1)

P₀ :Initial  linear momentum quantity

Pf : Initial  linear momentum quantity

Nomenclature and data

m₁: ball mass= 5.21 g= 5.21*10⁻³kg

V₀₁: initial ball speed,  =412 m/s

Vf₁: final ball speed

m₂: block mass =  14.8 kg  

V₀₂: initial block speed, = 0

Vf₂: final block speed

Problem development

For this problem the collision is perfectly elastic ,then, In addition to the linear moment, the kinetic energy is also conserved.

We assume that the ball moves to the right before the collision y continues moving  to the right after the collision .(+)

We assume that the block moves to the right before the collision.(+)

We apply furmula (1)

P₀=Pf

m₁*V₀₁+m₂*V₀₂=m₁*Vf₁+m₂*Vf₂

5.21*10⁻³*412+14.8*0= 5.21*10⁻³*Vf₁+14.8*Vf₂

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

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

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*10⁻³(Vf₂-412)+14.8*Vf₂

2.15= 5.21*10⁻³*Vf₂-2.15+14.8*Vf₂

4.3=14.805Vf₂

Vf₂ =4.3/14.805= 0.29 m/s : (+) ,with equal direction  of the movement of the ball before the collision.

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i already answered it by 34/s