A 2.0-kg ball moving eastward at 3.0 m/s suddenly collides with and sticks to a 4.0-kg ball moving northward at 2.0 m/s. What is the magnitude of the momentum of this system just after the collision?
A) $6.0 \mathrm{~kg}=\mathrm{m} / \mathrm{s}$
B) $8.0 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
C) $10 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
D) $14 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
E) $2.0 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$

Question

A  2.0-kg ball moving eastward at  3.0 m/s suddenly collides with and sticks to a  4.0-kg ball moving northward at  2.0 m/s. What is the magnitude of the momentum of this system just after the collision?
A) $6.0 \mathrm{~kg}=\mathrm{m} / \mathrm{s}$
B) $8.0 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
C) $10 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
D) $14 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
E) $2.0 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$

Quizplus · Accepted Answer

The total momentum of the system just after the collision can be found by calculating the vector sum of the momenta of the two balls before the collision. The momentum of the 2.0-kg ball is $2.0 \, \text{kg} \times 3.0 \, \text{m/s} = 6.0 \, \text{kg} \cdot \text{m/s}$ eastward, and the momentum of the 4.0-kg ball is $4.0 \, \text{kg} \times 2.0 \, \text{m/s} = 8.0 \, \text{kg} \cdot \text{m/s}$ northward. The magnitude of the total momentum is found using the Pythagorean theorem: $\sqrt{(6.0)^2 + (8.0)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, \text{kg} \cdot \text{m/s}$.

A 20-Kg Ball Moving Eastward at 3 6.0 kg=m/s6.0 \mathrm{~kg}=\mathrm{m} / \mathrm{s}6.0 kg=m/s

A 20-Kg Ball Moving Eastward at 3 $6.0 \mathrm{~kg}=\mathrm{m} / \mathrm{s}$