Momentum — Solutions
Full worked solutions, with method and final answer for every part.
Question 1
(a)p = mv, with mass converted to kg (58 g = 0.058 kg).
p = 0.058 × 45 = 2.61 kg·m/s
Question 2
(a)Total momentum before = total momentum after: m1u1 + m2u2 = (m1+m2)v.
(2.0 × 3.0) + (1.0 × 0) = (2.0+1.0) v → v = 6.0 ÷ 3.0 = 2.0 m/s
(b)Conservation of momentum: in a closed system with no external forces, total momentum before a collision equals total momentum after.
Total momentum is conserved
Question 3
(a)Take the initial direction as positive, so the rebound velocity is −6.0 m/s. Δp = m(v − u).
Δp = 0.16 × (−6.0 − 8.0) = 0.16 × (−14.0) = −2.24 kg·m/s
(b)F = Δp / Δt.
F = −2.24 ÷ 0.05 = −44.8 N, so the wall exerts about 44.8 N on the ball (directed away from the wall)
Question 4
(a)Total momentum before pushing off is zero (both at rest). By conservation of momentum, total momentum must still be zero afterward, so the skaters' momenta must be equal and opposite — meaning they move in opposite directions.
Momentum starts at zero and must stay at zero, so the two momenta cancel — equal and opposite motion
Question 5
(a)p = mv.
1200 × 20 = 24 000 kg·m/s
Question 6
(a)F = Δp/Δt = m(v−u)/t.
900 × (15−0) ÷ 3 = 4500 N
Question 7
(a)Total momentum before = total momentum after: m1u1 + m2u2 = (m1+m2)v, taking the 2kg trolley's velocity as −1 m/s.
(3×4) + (2×−1) = 5v → v = 10 ÷ 5 = 2 m/s
Question 8
(a)Total momentum starts at zero, so the astronaut's momentum must be equal and opposite to the tool's: m_astro × v_astro = −m_tool × v_tool.
v = (0.5×100) ÷ 50 = 1 m/s (in the opposite direction to the tool)
Question 9
(a)In a closed system with no external forces, the total momentum before an interaction equals the total momentum after it.
Total momentum is conserved in a closed system
Question 10
(a)Padding increases the time over which the impact momentum change happens. Since F = Δp/Δt, a longer contact time for the same change in momentum results in a smaller average force on the person hit.
Padding increases contact time, which reduces the average force for the same momentum change
Book a free trial class
Tell us a little about your requirement.