A 0.70 kg ball is moving horizontally at 5.8 m/s when it strikes a vertical wall and rebounds with speed 1.0 m/s. What is the magnitude of the change in its linear momentum? ____ kg • m/s.



Answer :

Answer:

Approximately [tex]4.8\; {\rm kg\cdot m\cdot s^{-1}}[/tex].

Explanation:

If an object of mass [tex]m[/tex] travels at a velocity of [tex]\vec{v}[/tex], the momentum of that object would be [tex]\vec{p} = m\, \vec{v}[/tex]. Note that momentum is a vector quantity, where the direction would be the same as the direction of the velocity.

If the original direction of motion of the ball is considered as the positive direction, the initial velocity of the ball would be positive: [tex]v_{0} = 5.8\; {\rm m\cdot s^{-1}}[/tex]. The new velocity of the ball, which is in the opposite direction, would be negative: [tex]v_{1} = (-1.0)\; {\rm m\cdot s^{-1}}[/tex].

To find the change in the momentum of the ball, subtract the initial value of momentum from the new value:

[tex]\begin{aligned} (\vec{p}_{1} - \vec{p}_{0}) &= m\, \vec{v}_{1} - m\, \vec{v}_{0} \\ &= (m)\, (\vec{v}_{1} - \vec{v}_{0}) \\ &= (0.70\; {\rm kg})\, ((-1.0)\; {\rm m\cdot s^{-1}} - 5.8\; {\rm m\cdot s^{-1}}) \\ &\approx (-4.76)\; {\rm kg\cdot m\cdot s^{-1}}\end{aligned}[/tex].

Therefore, the magnitude of the change in linear momentum would be approximately [tex]4.8\; {\rm kg\cdot m\cdot s^{-1}}[/tex] (rounded to two significant figures.)