Question

Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic collision. Momentum is conserved. Object A has a mass of mA = 16.1 kg and an initial velocity of = 8.11 m/s, due east. Object B, however, has a mass of mB = 29.8 kg and an initial velocity of = 5.46 m/s, due north. Find the (a) magnitude and (b) direction of the total momentum of the two-object system after the collision.

Answer 1

Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic collision. Momentum is conserved. Object A has a mass of mA = 16.1 kg and an initial velocity of = 8.11 m/s, due east. Object B, however, has a mass of mB = 29.8 kg and an initial velocity of = 5.46 m/s, due north. Find the (a) magnitude and (b) direction of the total momentum of the two-object system after the collision.

Solution

Momentum is conserved. That's why

$$\overrightarrow {P _ {A}} + \overrightarrow {P _ {A}} = \overrightarrow {P _ {A + B}},$$

where $\overrightarrow{P_A}$ is momentum of object A, $\overrightarrow{P_B}$ is momentum of object B, $\overrightarrow{P_{A + B}}$ is momentum of the two-object system after the collision.

Break each vector down into its horizontal and vertical components (y – due to north, x – due to east):

$\overrightarrow{P_A}$ is $(m_A v_A, 0)$, $\overrightarrow{P_B}$ is $(0, m_B v_B)$, $\overrightarrow{P_{A + B}}$ is $(m_A v_A, m_B v_B)$.

The magnitude of the total momentum of the two-object system after the collision:

$$\left| \overrightarrow {P _ {A + B}} \right| = \sqrt {(m _ {A} v _ {A}) ^ {2} + (m _ {B} v _ {B}) ^ {2}} = \sqrt {(1 6 . 1 * 8 . 1 1) ^ {2} + (2 9 . 8 * 5 . 4 6) ^ {2}} = 2 0 9 k g * \frac {m}{s}.$$

The direction of the total momentum of the two-object system after the collision:

$$\theta = \tan^ {- 1} \left(\frac {m _ {B} v _ {B}}{m _ {A} v _ {A}}\right) = \tan^ {- 1} \left(\frac {2 9 . 8 * 5 . 4 6}{1 6 . 1 * 8 . 1 1}\right) = 5 1 {}^ {\circ} \mathrm {n o r t h o f e a s t}.$$

Answer: $209\, kg * \frac{m}{s}$ at $51{}^{\circ}$ north of east.