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Marianna [84]
2 years ago
8

A 0.58 kg mass is moving horizontally with a speed of 6.0 m/s when it strikes a vertical wall. The mass rebounds with a speed of

4.0 m/s. What is the magnitude of the change in linear momentum of the mass
Physics
1 answer:
Sunny_sXe [5.5K]2 years ago
7 0

Answer:

5.8\; {\rm kg\cdot m \cdot s^{-1}}.

Explanation:

If the mass of an object is m and the velocity of that object is v, the linear momentum of that object would be m\, v.

Assume that the initial velocity of the mass is positive (6.0\; {\rm m\cdot s^{-1}}.) However, the direction of the velocity is reversed after the impact. Thus, the sign of the new velocity of the object would be negative- the opposite of that of the initial velocity. The new velocity would be (-4.0\; {\rm m\cdot s^{-1}}).

Thus, the change in the velocity of the mass would be:

\begin{aligned}& (\text{Change in Velocity}) \\ =\; & (\text{Final Velocity}) - (\text{Initial Velocity}) \\ =\; & (-4.0\; {\rm m\cdot s^{-1}}) - (6.0\; {\rm m\cdot s^{-1}}) \\ =\; & (-10\; {\rm m\cdot s^{-1})\end{aligned}.

The change in the linear momentum of the mass would be:

\begin{aligned} & \text{change in momentum} \\ =\; & (\text{mass}) \times (\text{change in velocity}) \\ =\; & 0.58\; {\rm kg} \times (-10\; {\rm m\cdot s^{-1}}) \\  =\; & (-5.8\; {\rm kg \cdot m \cdot s^{-1}})\end{aligned}.

Thus, the magnitude of the change of the linear momentum would be 5.8\; {\rm kg \cdot m \cdot s^{-1}}.

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