Question

A 0.10-kilogram piece of modeling clay is tossed

Answer 1

a) The initial speed of the clay is 30 m/s

b) The final velocity of the block increases (40 m/s)

Explanation:

a)

We can solve this problem by using the law of conservation of momentum. In fact, in absence of external forces (=no friction), the total momentum of the clay + block system is conserved before and after the collision. Therefore, we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v$  

where:  

$m_1 = 0.10 kg$ is the mass of the clay

$u_1$ is the initial velocity of the clay

$m_2 = 0.10 kg$ is the mass of the wood block

$u_2 = 0$ is the initial velocity of the wood block (at rest)

$v = 15 m/s$ is the final combined velocity of the clay+block after the collision

Re-arranging the equation, we can find $u_1$, the speed at which the clay was tossed:

$u_1 = \frac{(m_1+m_2)v-m_2 u_2}{m_1}=\frac{(0.10+0.10)(15)-0}{0.10}=30 m/s$

b)

In this second case, the clay is replaced by a bouncy ball, which rebounds back after the collision, instead of sticking with the block.

In this second case, the law of conservation of momentum becomes:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$  

where

$m_1 = 0.10 kg$ is the mass of the ball

$u_1 = 30 m/s$ is the initial velocity of the ball

$m_2 = 0.10 kg$ is the mass of the wood block

$u_2 = 0$ is the initial velocity of the wood block (at rest)

$v_1 = -10 m/s$ is the velocity of the bouncing ball after the collision (negative because it goes backward)

$v_2$ is the velocity of the block after the collision

Solving for $v_2$, we find the final velocity of the block:

$v_2 = \frac{m_1 u_1 - m_1 v_1}{m_2}=\frac{(0.10)(30)-(0.10)(-10)}{0.10}=40 m/s$

As we can see, the final velocity of the block has increased. The reason for that is that, as the ball bounces back, part of the total momentum is "carried away" by the ball in the backward direction, and since the total momentum must remain constant, this means that the momentum carried by the block in the forward direction must be larger than the previous situation.

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