Question

Find the impulse of a 50. kg object under the following scenarios:

a. The object accelerates to 7.5 m/s from rest.
b. The object stops from a velocity of 12.0 m/s.
C. The object change in velocity from 2.2 m/s to 6.3
d. The object hits the ground with 2.5 m/s and rebounds with the same speed

Answer 1

• a.
  `J = 375\, \rm N \cdot s`.
• b.
  `J = -600\; \rm N \cdot s`.
• c.
  `J = 205\; \rm N \cdot s`.
• d.
  `J = -250\; \rm N \cdot s`.

Step-by-step explanation:

When the velocity of an object changes, it would experience an impulse. If the mass of the object stays the same, and that the object moves along a line, the value of the impulse
`J` would be:

`J = m \cdot \Delta v`, where

• 
  `m` is the mass of the object, and
• 
  `\Delta v` is the change in the velocity of the object.

On the other hand, the change in the object's velocity can be found with the equation:

`\Delta v = v_{\text{final}} - v_\text{initial}`.

Note that if
`m` is in
`\rm kg` and
`\Delta v` is in
`\rm m \cdot s^(-1)`, the unit of
`J` would be
`\rm N \cdot s`.

a.

`v_\text{initial} = 0\; \rm m \cdot s^(-1)`.

`v_\text{final} = 7.5\; \rm m \cdot s^(-1)`.

`\Delta v = v_{\text{final}} - v_\text{initial} = 7.5\; \rm m \cdot s^(-1)`.

`J = m \cdot \Delta v = 50 * 7.5 = 375\; \rm N \cdot s`.

b.

`v_\text{initial} = 12.0\; \rm m \cdot s^(-1)`.

`v_\text{final} = 0\; \rm m \cdot s^(-1)`.

`\Delta v = v_{\text{final}} - v_\text{initial} = -12.0\; \rm m \cdot s^(-1)`

`J = m \cdot \Delta v = 50 * 12.0 = 600\; \rm N \cdot s`.

c.

`v_\text{initial} = 2.2\; \rm m \cdot s^(-1)`.

`v_\text{final} = 6.3\; \rm m \cdot s^(-1)`.

`\Delta v = v_{\text{final}} - v_\text{initial} = 4.1\; \rm m \cdot s^(-1)`.

`J = m \cdot \Delta v = 50 * 4.1 = 201\; \rm N \cdot s`.

d.

`v_\text{initial} = 2.5\; \rm m \cdot s^(-1)`.

`v_\text{final} = -2.5\; \rm m \cdot s^(-1)`.

Note that
`v_\text{final}` and
`v_\text{initial}` are of opposite signs. The reason is that the object's velocity has changed direction in this period.

`\Delta v = v_{\text{final}} - v_\text{initial} = -5.0\; \rm m \cdot s^(-1)`.

`J = m \cdot \Delta v = 50 * (-5.0) = -250\; \rm N \cdot s`.