Question

A 26.0 g ball is fired horizontally with initial speed v0 toward a 110 g ball that is hanging motionless from a 1.10 m -long string. The balls undergo a head-on, perfectly elastic collision, after which the 110 g ball swings out to a maximum angle θmax = 50.0. What was v0?


I think you need to find the tangential velocity using the angle that the ball swings to, but I am not sure how to go about beginning this problem.

Answer 1

Answer:

$7.3 ms^{-1}$

Explanation:

Consider the motion of the ball attached to string.

In triangle ABD

$Cos50 = \frac{AB}{AD} \\Cos50 = \frac{AB}{L}\\AB = L Cos50$

height gained by the ball is given as

$h = BC = AC - AD \\h = L - L Cos50\\h = 1.10 - 1.10 Cos50\\h = 0.393 m$

$M$  = mass of the ball attached to string = 110 g

$V$ = speed of the ball attached to string just after collision

Using conservation of energy

Potential energy gained = Kinetic energy lost

$Mgh = (0.5) M V^{2} \\V = sqrt(2gh)\\V = sqrt(2(9.8)(0.393))\\V = 2.8 ms^{-1}$

Consider the collision between the two balls

$m$  = mass of the ball fired = 26 g

$v_{o}$ = initial velocity of ball fired before collision = ?

$v_{f}$ = final velocity of ball fired after collision = ?

using conservation of momentum

$m v_{o} = MV + m v_{f}\\26 v_{o} = (110)(2.8) + 26 v_{f}\\v_{f} = v_{o} - 11.85$

Using conservation of kinetic energy

$m v_{o}^{2} = MV^{2} + m v_{f}^{2} \\26 v_{o}^{2} = 110 (2.8)^{2} + 26 (v_{o} - 11.85)^{2} \\v_{o} = 7.3 ms^{-1}$

Answer 2

At a maximum angle of 50°, the initial velocity ($V_0$) of the ball is equal to 7.3 m/s.

Given the following data:

Mass of ball 1 = 26.0 g.

Mass of ball 2 = 110.0 g.

Length = 1.10 m.

Maximum angle = 50°

How to calculate the initial velocity.

First of all, we would determine the height of the ball in motion through this derivation:

$h = L-Lcos\theta\\\\h = 1.10-1.10cos50\\\\h = 1.10-0.7071$

Height, h = 0.3929 meter.

Next, we would determine the velocity of the ball by applying the law of conservation of energy:

$P E=KE\\\\mgh=\frac{1}{2} mv^2\\\\V=\sqrt{2gh} \\\\V=\sqrt{2 \times 9.8 \times 0.3929 }$

V = 2.7750 m/s.

Also, we would determine the final velocity by applying the law of conservation of momentum:

$m_1v_o=m_1vf+m_2V\\\\26v_0=26v_f+110(2.7750)\\\\26v_f=26v_0-305.25\\\\v_f=(v_0-11.7404)\;m/s$

Now, we can determine the initial velocity:

$m_1v_o^2=m_1v_f^2+m_2V^2\\\\26v_0^2=26(v_0-11.7404)^2+110(2.7750)^2\\\\26v_0^2=26(v_0-11.7404)^2-847.0688\\\\V_0=7.3\;m/s$

Read more on kinetic energy here: brainly.com/question/1242059