a 0.70 kg ball is moving horizontally at 7.3 m/s when it strikes a vertical wall and rebounds with speed 2.2 m/s. what is the magnitude of the change in its linear momentum?

The magnitude of the change in the ball's linear momentum is equal to 6.65 kg · m/s.What is the linear momentum?Linear momentum, or simply called momentum, is defined to be mass times velocity. It measures how much mass in how much much motion. Momentum is mathematically expressed as p = mv where p is the momentum, m is the mass and v is the velocity.Momentum is a quantity that is conserved. Therefore, in a collision where momentum is involved, the initial momentum of an object must be equal to the final momentum of the object.To find the change in momentum of the .7-kg ball, we can use the formula Δp = m(u - v) where Δp is the change in momentum, m is the mass, u is the initial momentum and v is the final momentum.Δp = m(u - v)Δp = (0.7) (7.3 - (- 2.2))Δp = (0.7) (7.3 + 2.2)Δp = (0.7) (9.5)Δp = 6.65We have confirmed that the change in momentum of the ball is equal to 6.65 kg · m/s.Learn more about a similar problem of momentum here: https://brainly.com/question/25121535#SPJ4