A constant force of 20\text{N} acts on an object for 0.25\text{s}. What impulse does it deliver?

Answer: 5.0\text{N·s} J = F \Delta t = (20)(0.25) = 5.0\text{N·s}.

A 0.20\text{kg} ball at 5\text{m/s} is struck and rebounds at 5\text{m/s} (opposite direction). The impulse is

Answer: 2\text{N·s} J = \Delta p = m(vf - vi) = 0.20(5 - (-5)) = 2\text{N·s}.

Answer: 6\text{N·s} \Delta p = (0.5)(-6) - (0.5)(6) = -6\text{kg·m/s} (impulse magnitude 6\text{N·s} — twice the absorption case).

AP Physics 1 · Topic 4.2

Answer: 6\text{N·s} \Delta p = (0.5)(-6) - (0.5)(6) = -6\text{kg·m/s} (impulse magnitude 6\text{N·s} — twice the absorption case).

Part of Linear Momentum.(TOP-4.B)

Practice questions

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Sample 1difficulty 1/5
$20~\text{N}$
- A
  $5.0~\text{N·s}$
  check_circle
- B
  $500~\text{N·s}$
- C
  $0.80~\text{N·s}$
- D
  $80~\text{N·s}$
Why
$J = F \Delta t = (20)(0.25) = 5.0~\text{N·s}$
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Sample 2difficulty 1/5
$0.20~\text{kg}$
- A
  $0$
- B
  $10~\text{N·s}$
- C
  $1~\text{N·s}$
- D
  $2~\text{N·s}$
  check_circle
Why
$J = \Delta p = m(v_f - v_i) = 0.20(5 - (-5)) = 2~\text{N·s}$
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Sample 3difficulty 2/5
$6~\text{m/s}$
- A
  $0$
- B
  $12~\text{N·s}$
- C
  $6~\text{N·s}$
  check_circle
- D
  $3~\text{N·s}$
Why
$\Delta p = (0.5)(-6) - (0.5)(6) = -6~\text{kg·m/s}$
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Sample 4difficulty 2/5
$0.5~\text{kg}$
- A
  $1.5~\text{N·s}$
- B
  $6~\text{N·s}$
- C
  $0$
- D
  $3~\text{N·s}$
  check_circle
Why
$\Delta p = 0 - (0.5)(6) = -3~\text{kg·m/s}$
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Sample 5difficulty 2/5
$t = 2~\text{s}$
- A
  $4~\text{N·s}$
- B
  $12~\text{N·s}$
- C
  $16~\text{N·s}$
- D
  $8~\text{N·s}$
  check_circle
Why
$\Delta p = 12 - 4 = 8~\text{N·s}$
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