Question

Two balls of clay, with masses M1 = 0.49 kg and M2 = 0.47 kg, are thrown at each other and stick when they collide. Mass 1 has a velocity v1 = 3.2i m/s and mass 2 has a velocity of v2 = 1.6j m/s. Part (a) Write an expression for the initial momentum of the system. P, in terms of the masses M1 and M2, the magnitudes of the velocities vi and v2, and the unit vectors i and j Part (b) Find the horizontal component of the final velocity vx, in meters per second, after the collision Δ 33% Part (c) Find the vertical component of the final velocity vy, in meters per second, after the collision

Answer 1

Answer:

a) $p_i=1.568\hat{i}+0.752 \hat{j}$

b) $v_{fx}=1.668\ m.s^{-1}$

c) $v_{fy}=0.7999\ m.s^{-1}$

Explanation:

Given masses:

$m_1=0.49\ kg$

$m_2=0.47\ kg$

Velocity of mass 1, $v_1=3.2 \hat{i}\ m.s^{-1}$

Velocity of mass 2, $v_2=1.6 \hat{j}\ m.s^{-1}$

a)

Initial momentum:

$p_i=m_1.v_1+m_2.v_2$

$p_i=0.49\times 3.2 \hat{i}+0.47\times 1.6 \hat{j}$

$p_i=1.568\hat{i}+0.752 \hat{j}$

b)

magnitude of initial momentum:

$p_i=\sqrt{1.568^2+0.752 ^2}$

$p_i=1.739\ kg.m.s^{-1}$

From the conservation of momentum:

$p_f=p_i$

$m_f.v_f=1.739$

$v_f=\frac{1.739}{0.49+0.47}$

$v_f=1.85\ m.s^{-1}$ is the magnitude of final velocity.

Direction of final velocity will be in the direction of momentum:

$tan\theta=\frac{0.752 }{1.568}$

$\theta=25.62^{\circ}$

$\therefore v_{fx}=1.85\ cos25.62^{\circ}$

$v_{fx}=1.668\ m.s^{-1}$

c)

Vertical component of final velocity:

$v_{fy}=1.85\ sin 25.62^{\circ}$

$v_{fy}=0.7999\ m.s^{-1}$