Question

A 3.9 g dart is fired into a block of wood with a mass of 24.6 g. The wood block is initially at rest on a 1.5 m tall post. After the collision, the wood block and dart land 3.5 m from the base of the post. Find the initial speed of the dart.

Answer 1

Answer:

46.48m/s

Explanation:

The problem is a combination of the principle of conservation of linear momentum and projectile motion.

The principle of conservation of linear momentum states that in a closed system, the total momentum of colliding bodies before impact is equal to the total momentum after impact. The masses stated in the problem experienced an inelastic collision. In an inelastic collision, the bodies involved stick together after the collision and move with a common velocity.

For two bodies of masses $m_1$ and $m_2$ moving with velocities $u_1$ and $u_2$ before impact, if they experience inelastic collision, the conservation of their momenta is as stated in equation (1);

$m_1u_1+m_2u_2=(m_1+m_2)v..................(1)$

were v is their common velocity after impact. If the second mass $m_2$ was at rest before the impact, then its initial velocity $u_2=0m/s$. therefore $m_2u_2=0$. Equation (1) then becomes;

$m_1u_1=(m_1+m_2)v..............(2)$

In the problem stated, the second mass taken as the mass of the wooden block was at rest before the impact and the collision was inelastic since both the wood and the dart stuck together and moved with a common velocity after the impact. Therefore we can use equation (2) for the problem.

Given;

$m_1=3.9g=0.0039kg\\u_1=?\\m_2=24.6g=0.0246kg\\v=?$

Substituting these values into (2), we get the following;

$0.0039*u_1=(0.0039+0.024)v\\0.0039u_1=0.0285v.........(3)$

Their common v velocity after impact now makes both the wooden block and the dart (as a single body) to fall vertically through a height h of 1.5m over a range R of 3.5m as stated by the problem; hence by the principle of projectile motion for a body projected horizontally, the following relationship holds;

$R= vt............(4)$

were t is the time taken to fall through the height h. To obtain t we use the second equation of free fall under gravity;

$h=\frac{1}{2}gt^2...........(5)$

were g is acceleration due to gravity taken as $9.8m/s^2$. Therefore;

$1.5=\frac{1}{2}*9.8*t^2\\1.5=4.9t^2\\t^2=\frac{1.5}{4.9}=0.306\\t=\sqrt{0.306} =0.55s$

We then substitute R and t into equation (4) to obtain v.

$3.5=v*0.55\\v=\frac{3.5}{0.55}\\v=6.36m/s$

We now further substitute this value of v into (3) to obtain $u_1$;

$u_1=\frac{0.0285v}{0.0039}\\\\u_1=\frac{0.0285*6.36}{0.0039}\\\\u_1=\frac{0.18126}{0.0039}\\\\u_1=46.48m/s$