# Impulse Momentum Exam 2 and Problem Solutions

**1.** Objects shown in the figure collide and stick and move together. Find final velocity objects.
Using conservation of momentum law;

**m₁.V₁+m₂.V₂=(m₁+m₁).Vfinal**

**3.8+4.10=7.Vfinal**

**64=7.Vfinal**

**Vfinal=9,14m/s**

**2.** 2kg and 3kg objects slide together, and then they break apart. If the final velocity of m₂ is 10 m/s,

**a)** Find the velocity of object m₁.

**b)** Find the total change in the kinetic energies of the objects.
**a)** Using conservation of momentum law;

**(m₁+m₂).V=m₁.V₁+m₂.V₂**

**5.4=30+2.V₁**

**V₁=-5m/s**

**b)** **EKinitial=1/2/m₁+m₂).V₂**

**EKinitial=1/2.5.16=40joule**

**EKfinal=1/2.2.52+1/2.3.102**

**EKfinal=175 joule**

**Change in the kinetic energy is =175-40=135 joule**

**3.** As shown in the figure below, object m₁ collides stationary object m₂. Find the magnitudes of velocities of the objects after collision. (elastic collision)
In elastic collisions we find velocities of objects after collision with following formulas;

**V₁’=(m₁-m₂)/(m₁+m₂).V₁**

**V₂’=(2m₁/m₁+m₂).V₁**

**m₁=6kg, m₂=4kg, V₁=10m/s**

**V₁’=(6-4/6+4).10=2m/s**

**V₂’=(2.6/6+4).10=12m/s**

**4.** Momentum vs. time graph of object is given below. Find forces applied on object for each interval.
**F.Δt=ΔP**

**F=ΔP/Δt**

Slope of the graph gives us applied force.

**I. Interval:**

**F₁=P₂-P₁/10-0=-50/10=-5N**

**II. Interval:**

**F₂=50-50/10=0**

**III. Interval:**

**F₃=100-50/10=5N**

**5.** A box having mass 0,5kg is placed in front of a 20 cm compressed spring. When the spring released, box having mass m₁, collide box having mass m₂ and they move together. Find the velocity of boxes.
Energy stored in the spring is transferred to the object m1.

**1/2.k.X²=1/2.mV²**

**50N/m.(0,2)²=0,5.V²**

**V=2m/s**

Two object do inelastic collision.

**m₁.V₁=(m₁+m₂).Vfinal**

**0,5.2=2.Vfinal**

**Vfinal=0,5m/s**