Question

question 6.2a:
two blocks slide on a collision course across a frictionless surface, as in the figure. the resulting collision is inelastic. the first block has mass ( m = 1.00 mathrm{~kg} ) and is initially sliding due north at a speed of ( v_{1}=8.95 mathrm{~m} / mathrm{s} ). the second block has mass ( m = 7.40 \times 10^{-2} mathrm{~kg} ) and is initially sliding at a speed of ( v_{i}=13.5 mathrm{~m} / mathrm{s} ) directed at an angle ( \theta = 29.5^{circ} ) south of east. immediately after the inelastic collision, the second block is observed sliding at ( v_{f}=4.95 mathrm{~m} / mathrm{s} ) in a direction of ( varphi = 17.5^{circ} ) north of east.

determine the components of the first blocks velocity after the collision.
( v_{x f}=quad mathrm{m} / mathrm{s} )
( v_{y f}=quad mathrm{m} / mathrm{s} )
question 6.2b:

Explanation:

Step1: Apply conservation of momentum in x - direction

The initial momentum in the x - direction is \(p_{ix}=mv_{i}\cos\theta\) (only block 2 has initial x - momentum). The final momentum in the x - direction is \(p_{fx}=mv_{f}\cos\varphi+MV_{xf}\).
By conservation of momentum \(p_{ix} = p_{fx}\), so \(mv_{i}\cos\theta=mv_{f}\cos\varphi+MV_{xf}\).
Substitute \(M = 1.00\space kg\), \(m = 7.40\times10^{-2}\space kg\), \(v_{i}=13.5\space m/s\), \(\theta = 29.5^{\circ}\), \(v_{f}=4.95\space m/s\), \(\varphi = 17.5^{\circ}\)

$$
LATEXBLOCK0
$$

Step2: Apply conservation of momentum in y - direction

The initial momentum in the y - direction is \(p_{iy}=MV_{i}-mv_{i}\sin\theta\) (block 1 has positive y - momentum and block 2 has negative y - momentum). The final momentum in the y - direction is \(p_{fy}=mv_{f}\sin\varphi+MV_{yf}\).
By conservation of momentum \(p_{iy}=p_{fy}\), so \(MV_{i}-mv_{i}\sin\theta=mv_{f}\sin\varphi+MV_{yf}\)
Substitute the values:

$$
LATEXBLOCK1
$$

Answer:

\(V_{xf}=0.521\space m/s\), \(V_{yf}=8.35\space m/s\)