Question

an object of mass 1 kg travelling in a straight line with a velocity of 10 metre per second collides with and sticks to a stationary block of mass 5 kg then they both move off together in the same straight line calculate the momentum just before the impact and just after the impact also calculate the velocity of the combined object​

Answer 1

Explanation:

Mass of the object, m1 = 1 kg

Velocity of the object before collision, v1 = 10 m/s

Mass of the stationary wooden block, m2 = 5 kg

Velocity of the wooden block before collision, v2 = 0 m/s

ˆ´ Total momentum before collision = m1 v1 + m2 v2

= 1 (10) + 5 (0) = 10 kg m sˆ’1

It is given that after collision, the object and the wooden block stick together.

Total mass of the combined system = m1 + m2

Velocity of the combined object = v

According to the law of conservation of momentum:

Total momentum before collision = Total momentum after collision

m1 v1 + m2 v2 = (m1 + m2) v

1 (10) + 5 (0) = (1 + 5) v

v = 10 / 6

= 5 / 3

The total momentum after collision is also 10 kg m/s.

Total momentum just before the impact = 10 kg m sˆ’1

Total momentum just after the impact = (m1 + m2) v = 6 — 5 / 3 = 10 kg ms-1

Hence, velocity of the combined object after collision = 5 / 3 ms-1

Answer 2

Explanation :-

Given :

$\sf{}Mass\ of\ the\ first\ object,m_1= 1kg$

$\sf{}Mass\ of\ the\ second\ object,m_2 = 5kg$

$\sf{}Initial\ velocity\ of\ the\ first\ object,u_1 =10m/s$

$\sf{}Initial\ velocity\ of\ the\ second\ object,u_2 =0m/s$

To Find :

$\sf{}Total\ mometum\ before\ collision\ and\ after\ collision=\ ?$

$\sf{}\sf{}Final\ velocity\ of\ the\ combined\ object,v=\ ?$

Solution :

According to the conversation law of momentum,

$\sf{}\Rightarrow m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$\sf{}\Rightarrow 1kg\times 10kg+ 5kg\times 0m/s=1kg\times v +5\times 0$

$\sf{}\Rightarrow 1kg\times 10kg\ m/s=(1kg+5kg)v$

$\sf{}\Rightarrow 10kg\ m/s=6kg\times v$

$\sf{}\Rightarrow \dfrac{10kg\ m/s}{6kg}=v$

$\sf{}\Rightarrow \dfrac{5kg\ m/s}{3kg}=v$

$\sf{}\Rightarrow v=1.66m/s............(1)$

Hence,velocity is equal to 1.66m/s

Total Initial momentum,

$\sf{}m_1u_1+m_2u_2$

$\sf{}\Rightarrow 1kg \times 10m/s+5kg\times0 m/s$

$\sf{}\Rightarrow 10kg\ m/s$

Total momentum just after collision,

$\sf{} m_1v_1+m_2v_2$

$\sf{}\Rightarrow (m_1+m_2)v$

(Since,both the objects move with same velocity “v” after collision)

$\sf{}\Rightarrow (1kg+5kg)\times \dfrac{10}{6}m/s \ \ \ \ \ \ \(From\ equation\ (1)$

$\sf{}\Rightarrow 6kg \times \dfrac{10}{6}m/s$

$\sf{}\Rightarrow 10kg\ m/s$