A 2.0 kg ball traveling at 1.0 m/s strikes a 3.0 kg ball which is at rest. After an elastic collision, the 2.0 kg ball is traveling in a direction 45 degrees left of its original path. Set up equations that could be solved to find the velocity of the first ball and the speed and direction of the second ball.

Calculate the velocity of the target ball after the collision. Express your answer to two significant figures and include the appropriate units.   Calculate the mass of the target ball. Express your answer to two significant figures and include the appropriate units.

Ball 1 has a mass of 2.0 kg and is suspended with a 3.0 m rope from a post so that the ball is stationary. Ball 2 has a mass of 4.0 kg and is tied to another rope. The second rope also measures 3.0 m but is held at a 60.00 angle, as shown in Figure below. When ball 2 is released, it collides head-on, with ball 1 in an elastic collision. 3.0 m 60.0⁰ 3.0 m 2 4.0 kg 1 2.0 kg So far Ek = Eg 1/2 m2 V₁2² = m2gAy Vf₁ = (m₁-m₂/m₁ + m2)vi1 + (2m2/m1 + m2) V₁2 Vf2 = (m₂-mi/m₁ + m2)vi2 + (2m₁ /m₁ + m2) vil For the second part, you need to calculate the Eg₁ = Eki and Eg2 = Ek2, substitute the value of h and find the h maximum for 1 and 2 balls.

A 6.64 kg object is initially traveling in the positive x-direction at 9.44 m/s. It collides with a 7.85 kg object that is initially at rest. The two objects are in contact for just 14.7 ms. After the collision the 7.85 kg object is observed to move in the positive x-direction at 7.95 m/s. What is the average net force that acts on the 7.85 kg object during the 14.7 ms?